๐Ÿ“–Topic Explanations

๐ŸŒ Overview
Hello students! Welcome to the fascinating and fundamental topic of Differentiability!

Get ready to unlock a powerful concept that helps us understand the true nature of change and smoothness in the world of functions. This isn't just a chapter; it's a vital lens through which we analyze how mathematical relationships evolve!

Have you ever wondered why some curves are perfectly smooth, like a gentle wave, while others have sharp turns or breaks, like a zigzag path? Differentiability provides the mathematical framework to distinguish between these behaviors. At its core, Differentiability asks a profound question: "Can we find a unique, well-defined tangent line at every point on a function's curve?" If the answer is yes, the function is considered 'smooth' or differentiable at that point.

Think of it like driving a car. If the road is perfectly smooth, you can always determine your exact direction (tangent) at any instant. But if the road suddenly has a sharp turn or a huge pothole, your direction becomes ambiguous or undefined at that specific point. Functions behave in a remarkably similar fashion!

Understanding Differentiability is not just an academic exercise; it's a cornerstone of Calculus and has immense practical applications. From predicting the trajectory of a rocket to optimizing economic models, and from designing efficient algorithms to understanding the rates of chemical reactions, the principles of differentiability are at play.

For your JEE Main and Board examinations, Differentiability is a high-yield topic that often interlinks with Limits and Continuity. A solid grasp here will not only boost your scores but also build a robust foundation for advanced topics like applications of derivatives, integrals, and differential equations.

In this section, we will embark on a journey to:

  • Understand the precise definition of differentiability at a point and over an interval.

  • Explore its strong connection with continuity โ€“ are all continuous functions differentiable? The answer might surprise you!

  • Unravel the geometric interpretation of differentiability and what it means for a curve to be 'smooth'.

  • Identify the common scenarios where functions fail to be differentiable (think sharp corners, cusps, or vertical tangents).



Prepare to sharpen your analytical skills and gain a profound insight into the dynamics of mathematical functions. By mastering Differentiability, you'll be equipped with an essential tool for advanced problem-solving and a deeper appreciation for the elegance of calculus. Let's dive in!
๐Ÿ“š Fundamentals
Hello aspiring mathematicians! Welcome to this fundamental session on one of the most exciting and crucial topics in calculus: Differentiability. You might have heard of terms like "derivatives" or "differentiation" before, but today, we're going to dive deep into what it truly means for a function to be *differentiable* and why it's such a big deal.

Think of it like this: if continuity was about whether you could draw a function's graph without lifting your pen, differentiability is about whether that graph is "smooth" enough at every point, or if it has any sudden jolts, sharp corners, or vertical climbs.

Let's start our journey!

### 1. The Idea of "Slope" โ€“ A Quick Recap

You're probably familiar with the concept of the slope of a straight line, right? It's that wonderful measure of how steep a line is. For a straight line, the slope is constant everywhere. For example, if you're driving on a perfectly straight highway that goes uphill, the steepness (slope) is the same no matter where you are on that road. We calculate it as $frac{ ext{change in y}}{ ext{change in x}}$ or $frac{y_2 - y_1}{x_2 - x_1}$.

But what about a curve? A curve isn't straight; its steepness changes at every single point! Imagine driving on a winding, hilly road. Sometimes it's a gentle slope, sometimes it's super steep, sometimes it's flat, and sometimes it's going downhill. The "steepness" is constantly changing.

This changing steepness is where the concept of a tangent line comes in, and that's precisely what differentiability helps us define.

### 2. From Secant to Tangent: The Heart of the Idea

Let's take a function $y = f(x)$ and look at its graph.


Secant to Tangent Animation


(Imagine a curve, not a straight line, for our discussion.)

1. The Secant Line: Pick two distinct points on the curve, say $P(x, f(x))$ and $Q(x+h, f(x+h))$, where $h$ is a small change in $x$. If we draw a line connecting these two points, it's called a secant line. The slope of this secant line is:
$$m_{ ext{secant}} = frac{f(x+h) - f(x)}{(x+h) - x} = frac{f(x+h) - f(x)}{h}$$
This represents the average rate of change of $f(x)$ between $x$ and $x+h$.

2. Getting Closer and Closer: Now, what happens if we make point $Q$ get closer and closer to point $P$? Imagine $h$ becoming smaller and smaller, approaching zero. As $Q$ approaches $P$, the secant line starts to rotate, and it gets closer and closer to becoming a very special line: the tangent line at point $P$.

3. The Tangent Line: A tangent line is a line that "just touches" the curve at a single point, $P$, and has the same direction or steepness as the curve at that exact point. It represents the instantaneous rate of change at that specific point.

To find the slope of this tangent line, we use the magic of limits!

### 3. The Derivative: Our Instantaneous Slope Finder

The slope of the tangent line at a point $(x, f(x))$ on the curve $y=f(x)$ is called the derivative of $f(x)$ with respect to $x$ at that point. It's denoted by $f'(x)$ or $frac{dy}{dx}$ or $y'$.

Mathematically, we define it as:
$$f'(x) = lim_{h o 0} frac{f(x+h) - f(x)}{h}$$

This is the First Principle of Differentiation, and it's super important!
* The expression $frac{f(x+h) - f(x)}{h}$ is the slope of the secant.
* Taking the limit as $h o 0$ transforms it into the slope of the tangent.

Think of it this way: If you're checking the speed of a car. Average speed over an hour might be 60 km/h. But your instantaneous speed at a particular second could be 80 km/h if you just accelerated. The derivative gives you that "instantaneous speed" or "instantaneous rate of change" of the function at a precise point.

### 4. What Does Differentiability Actually Mean?

A function $f(x)$ is said to be differentiable at a point $x=a$ if the limit defining the derivative at that point exists and is a finite value.
That is, if:
$$f'(a) = lim_{h o 0} frac{f(a+h) - f(a)}{h}$$
exists and is a finite number.

If this limit exists, it means a unique, non-vertical tangent line can be drawn to the curve at $x=a$.

If a function is differentiable at *every* point in an interval, we say it is differentiable over that interval.

### 5. When Does a Function Fail to Be Differentiable? (The Nasty Cases)

For the derivative to exist, the limit $lim_{h o 0} frac{f(a+h) - f(a)}{h}$ must exist. This means that the "slope from the left" must equal the "slope from the right" at that point, and both must be finite. Let's break down the conditions where this might fail:

#### Condition 1: Discontinuity
A function MUST be continuous at a point to be differentiable at that point.
Why? Imagine a jump or a break in the graph. How can you draw a unique tangent line at a point where the function itself isn't even "connected"? You can't!
So, if $f(x)$ is discontinuous at $x=a$, then $f(x)$ is not differentiable at $x=a$.

Important take-away:





















Statement Truth Value Explanation
If a function is differentiable at $x=a$, then it is continuous at $x=a$. TRUE Differentiability is a stronger condition than continuity. It implies continuity.
If a function is continuous at $x=a$, then it is differentiable at $x=a$. FALSE Continuity is necessary, but not sufficient. There are continuous functions that are not differentiable.


This is a favourite concept for JEE questions!

#### Condition 2: Sharp Corners or Cusps (The "Pointy" Bits)
Even if a function is continuous, it might not be differentiable. The classic example is $f(x) = |x|$ at $x=0$.
The graph of $|x|$ looks like a "V". At $x=0$, it has a sharp corner.
If you try to draw a tangent line at $x=0$, what happens?
* If you approach $x=0$ from the left, the slope is -1.
* If you approach $x=0$ from the right, the slope is +1.
You can't draw a single, unique tangent line at $x=0$ that captures the "direction" of the curve there. It's like having two different roads meeting at a point, and you can't decide which way the "road" itself is going at that precise junction.

In calculus terms, this means the Left-Hand Derivative (LHD) and the Right-Hand Derivative (RHD) are not equal.

Definition of LHD and RHD:
* Left-Hand Derivative (LHD) at $x=a$:
$$f'(a^-) = lim_{h o 0^-} frac{f(a+h) - f(a)}{h} = lim_{x o a^-} frac{f(x) - f(a)}{x-a}$$
(This is the slope of the tangent approached from the left side of $a$)

* Right-Hand Derivative (RHD) at $x=a$:
$$f'(a^+) = lim_{h o 0^+} frac{f(a+h) - f(a)}{h} = lim_{x o a^+} frac{f(x) - f(a)}{x-a}$$
(This is the slope of the tangent approached from the right side of $a$)

For $f(x)$ to be differentiable at $x=a$, we MUST have: LHD = RHD = finite value.

For $f(x)=|x|$ at $x=0$:
* LHD at $x=0$: $lim_{h o 0^-} frac{|0+h| - |0|}{h} = lim_{h o 0^-} frac{-h}{h} = -1$ (since $h < 0$, $|h| = -h$)
* RHD at $x=0$: $lim_{h o 0^+} frac{|0+h| - |0|}{h} = lim_{h o 0^+} frac{h}{h} = 1$ (since $h > 0$, $|h| = h$)
Since LHD $
eq$ RHD, $f(x)=|x|$ is not differentiable at $x=0$.

#### Condition 3: Vertical Tangent
Sometimes, a function can be continuous, but at a certain point, the tangent line might be perfectly vertical.
Example: $f(x) = x^{1/3}$ (cube root of $x$) at $x=0$.
The derivative at $x=0$ involves $lim_{h o 0} frac{(0+h)^{1/3} - 0^{1/3}}{h} = lim_{h o 0} frac{h^{1/3}}{h} = lim_{h o 0} frac{1}{h^{2/3}}$.
As $h o 0$, $h^{2/3} o 0^+$, so $frac{1}{h^{2/3}} o infty$.
A vertical tangent means the slope is infinite, which means the derivative does not exist (it's not a finite value). So, the function is not differentiable at $x=0$.

### 6. Summary of Non-Differentiability Conditions

A function $f(x)$ is NOT differentiable at $x=a$ if any of these conditions are met:
1. Discontinuity: $f(x)$ is not continuous at $x=a$. (Jump, hole, or asymptote)
2. Sharp Corner / Cusp: LHD $
eq$ RHD at $x=a$. (Like $f(x)=|x|$ at $x=0$)
3. Vertical Tangent: LHD or RHD (or both) approach $pm infty$ at $x=a$. (Like $f(x)=x^{1/3}$ at $x=0$)

### 7. Why Is Differentiability Important?

Differentiability isn't just a theoretical concept; it's extremely practical!
* In Physics: The derivative of position gives velocity, and the derivative of velocity gives acceleration. If position isn't differentiable, you can't define instantaneous velocity!
* In Economics: Marginal cost or marginal revenue are derivatives.
* In Optimization: To find maximum or minimum values of a function, we often set its derivative to zero. This only works if the function is differentiable at those points!
* In Engineering and Computer Graphics: Smooth curves are essential. Differentiability helps ensure that curves transition seamlessly without awkward breaks or sharp edges.

### 8. CBSE vs. JEE Focus

* For CBSE Board exams (Class 12), the emphasis is often on applying differentiation rules (like product rule, quotient rule, chain rule) to find derivatives of various functions and solving problems based on rates of change, tangent/normal equations, and maxima/minima. Understanding the basic definition and continuity as a prerequisite is usually covered.
* For JEE Mains & Advanced, the concept of differentiability itself is a hotbed for questions. You'll often encounter problems asking to determine if a piecewise function is differentiable at a specific point, finding values of constants that make a function differentiable, or analyzing graphs for points of non-differentiability. The definitions involving LHD and RHD are crucial for these types of problems. You need to be able to apply the first principle definition rigorously.

JEE Alert: Always be on the lookout for functions defined piecewise, or involving absolute values, greatest integer functions, or fractional parts โ€“ these are common traps for non-differentiability!

Understanding differentiability from these fundamental principles will set a very strong foundation for your journey through calculus. Keep practicing, and you'll master it in no time!
๐Ÿ”ฌ Deep Dive
Welcome to a deep dive into the fascinating world of Differentiability! In the realm of Calculus, differentiability is a cornerstone concept that allows us to understand the "smoothness" and "rate of change" of functions. While you might have encountered derivatives as a formulaic calculation, here we'll rigorously define what it truly means for a function to be differentiable at a point, explore its profound geometric and physical interpretations, and understand its crucial relationship with continuity.

Let's begin our journey!

### 1. Introduction to Differentiability: What Does It Mean?

Remember our previous discussions on the derivative? We defined the derivative of a function `f(x)` at a point `x=a` as the instantaneous rate of change of `f(x)` with respect to `x` at that very point. It's like asking: how fast is your car's speed changing *right now*, at this exact moment?

Mathematically, this instantaneous rate of change at `x=a` is given by the limit:

$$f'(a) = lim_{h o 0} frac{f(a+h) - f(a)}{h}$$

Or, equivalently:

$$f'(a) = lim_{x o a} frac{f(x) - f(a)}{x - a}$$

A function `f(x)` is said to be differentiable at a point `x=a` if this limit exists and is a finite real number. If this limit does not exist, or if it's infinite, then the function is not differentiable at `x=a`.

Think of it this way: a differentiable function is "smooth" and "well-behaved" at that point. There are no sudden breaks, no sharp corners, and no vertical rises or falls.

### 2. Formal Definition: Left Hand and Right Hand Derivatives

For the limit to exist, the Left Hand Limit (LHL) must be equal to the Right Hand Limit (RHL). Applying this to our derivative definition, we introduce the concepts of Left Hand Derivative (LHD) and Right Hand Derivative (RHD).

At a point `x = a`:

1. Left Hand Derivative (LHD):
This represents the slope of the tangent approaching `x=a` from the left side.
$$LHD = f'(a^-) = lim_{h o 0^-} frac{f(a+h) - f(a)}{h} = lim_{x o a^-} frac{f(x) - f(a)}{x - a}$$
(Here, `h` approaches 0 from the negative side, or `x` approaches `a` from values less than `a`).

2. Right Hand Derivative (RHD):
This represents the slope of the tangent approaching `x=a` from the right side.
$$RHD = f'(a^+) = lim_{h o 0^+} frac{f(a+h) - f(a)}{h} = lim_{x o a^+} frac{f(x) - f(a)}{x - a}$$
(Here, `h` approaches 0 from the positive side, or `x` approaches `a` from values greater than `a`).

A function `f(x)` is differentiable at `x=a` if and only if:

  1. The LHD at `x=a` exists and is finite.

  2. The RHD at `x=a` exists and is finite.

  3. LHD = RHD


If all these conditions are met, then `f'(a) = LHD = RHD`.

### 3. Geometric Interpretation: The Slope of the Tangent

The most intuitive way to visualize differentiability is through geometry. The derivative `f'(a)` at a point `(a, f(a))` represents the slope of the tangent line to the curve `y = f(x)` at that point.

* If a function is differentiable at `x=a`, it means that there is a unique, non-vertical tangent line that can be drawn to the curve `y=f(x)` at the point `(a, f(a))`. The slope of this tangent line is precisely `f'(a)`.

* What happens if a function is not differentiable?

  • Sharp Corners or Cusps: Imagine the graph of `y = |x|` at `x=0`. If you try to draw a tangent line at the origin, you'll find that you can draw infinitely many lines that "touch" the graph at that point. The slope approaching from the left is -1, and from the right is +1. Since LHD โ‰  RHD, there isn't a unique tangent line, hence it's not differentiable. These points are often called "sharp corners" or "cusps."

  • Vertical Tangents: Consider `y = x^(1/3)` at `x=0`. If you try to find `f'(0)`, you'll find the limit tends to infinity. Geometrically, this means the tangent line at `x=0` is vertical. A vertical line has an undefined slope, so the derivative does not exist (it's not a finite real number).

  • Discontinuities: If a function has a jump, a hole, or an asymptote at a point, you cannot even connect the curve in a smooth way to draw a tangent line. We'll delve into this relationship next.



### 4. Physical Interpretation: Instantaneous Rate of Change

Beyond geometry, differentiability has profound implications in physics and other sciences:

* If `s(t)` represents the position of an object at time `t`, then `s'(t)` (the derivative) represents its instantaneous velocity at time `t`. For the object to have a well-defined instantaneous velocity, its position function must be differentiable. A sharp corner in a position-time graph would imply an instantaneous change in direction without passing through zero velocity, which is physically impossible.
* Similarly, if `v(t)` represents velocity, `v'(t)` represents instantaneous acceleration.
* In economics, the derivative of a cost function gives the marginal cost. For these concepts to be well-defined, the underlying functions must be differentiable.

### 5. The Crucial Link: Differentiability and Continuity

One of the most important theorems in calculus connects differentiability and continuity:

Theorem: If a function `f(x)` is differentiable at a point `x=a`, then it must be continuous at `x=a`.

Proof:
To prove that `f(x)` is continuous at `x=a`, we need to show that `$lim_{x o a} f(x) = f(a)$`, or equivalently, `$lim_{x o a} [f(x) - f(a)] = 0$`.

We know that if `f(x)` is differentiable at `x=a`, then `$lim_{x o a} frac{f(x) - f(a)}{x - a} = f'(a)$` exists and is finite.

Consider the expression `f(x) - f(a)`. For `x โ‰  a`, we can write:
$$f(x) - f(a) = left( frac{f(x) - f(a)}{x - a}
ight) cdot (x - a)$$

Now, let's take the limit as `x` approaches `a` on both sides:
$$lim_{x o a} [f(x) - f(a)] = lim_{x o a} left[ left( frac{f(x) - f(a)}{x - a}
ight) cdot (x - a)
ight]$$

Using the limit properties (limit of a product is product of limits, provided they exist):
$$lim_{x o a} [f(x) - f(a)] = left( lim_{x o a} frac{f(x) - f(a)}{x - a}
ight) cdot left( lim_{x o a} (x - a)
ight)$$

We know that `$lim_{x o a} frac{f(x) - f(a)}{x - a} = f'(a)$` (since `f` is differentiable).
And `$lim_{x o a} (x - a) = a - a = 0$`.

So,
$$lim_{x o a} [f(x) - f(a)] = f'(a) cdot 0$$
$$lim_{x o a} [f(x) - f(a)] = 0$$

This implies `$lim_{x o a} f(x) - lim_{x o a} f(a) = 0$`
Since `f(a)` is a constant with respect to `x`, `$lim_{x o a} f(a) = f(a)$`.
Thus, `$lim_{x o a} f(x) - f(a) = 0$`
Which means `$lim_{x o a} f(x) = f(a)$`.

This is the definition of continuity at `x=a`. Hence, the theorem is proven.

Important Converse:
The converse of this theorem is NOT true. A function can be continuous at a point but not differentiable at that point.
The classic example is `f(x) = |x|` at `x=0`.
* It is continuous at `x=0` because `$lim_{x o 0} |x| = 0 = f(0)$`.
* However, it is not differentiable at `x=0` because its graph has a sharp corner (LHD = -1, RHD = 1).

Key Takeaway: Differentiability is a stronger condition than continuity. If a function is differentiable, it's automatically continuous. But if it's continuous, it's not necessarily differentiable.


























Condition Implication Example
Differentiable at `a` Continuous at `a` `f(x) = x^2` at `x=0`
Continuous at `a` May or may not be differentiable at `a` `f(x) = |x|` at `x=0` (Continuous but not differentiable)
Not continuous at `a` Not differentiable at `a` `f(x) = [x]` (greatest integer function) at `x=1`


### 6. When Does Differentiability Fail? Common Scenarios

Based on our discussions, a function `f(x)` is NOT differentiable at `x=a` if:

1. `f(x)` is discontinuous at `x=a`: (e.g., jump, hole, vertical asymptote).
2. `f(x)` has a sharp corner or cusp at `x=a`: (LHD โ‰  RHD, but both are finite). E.g., `|x|`, `|x-2|`, `|sin x|` at `x=nฯ€`.
3. `f(x)` has a vertical tangent at `x=a`: (LHD or RHD, or both, tend to `ยฑโˆž`). E.g., `x^(1/3)` at `x=0`.

### 7. Differentiability of Piecewise Functions (JEE Focus!)

Piecewise functions are a favorite for JEE examiners when testing differentiability. For a function defined in pieces, you need to be particularly careful at the "joining points" โ€“ the values of `x` where the function definition changes.

Strategy for Piecewise Functions:
Let `f(x)` be a piecewise function defined as:
`f(x) = { g(x) for x โ‰ค a; h(x) for x > a }`

To check differentiability at `x = a`:

1. Check for Continuity at `x=a` FIRST!
* If `f(x)` is not continuous at `x=a`, then it is definitely not differentiable at `x=a`.
* For continuity, `$lim_{x o a^-} f(x) = lim_{x o a^+} f(x) = f(a)$`. This means `g(a) = h(a)`. If this condition fails, stop; not differentiable.

2. If Continuous, Check LHD and RHD:
* LHD at `x=a`: Calculate `g'(a)` (if `g(x)` is differentiable at `a`).
Alternatively, use the limit definition: `$lim_{h o 0^-} frac{f(a+h) - f(a)}{h} = lim_{h o 0^-} frac{g(a+h) - g(a)}{h}$`.
* RHD at `x=a`: Calculate `h'(a)` (if `h(x)` is differentiable at `a`).
Alternatively, use the limit definition: `$lim_{h o 0^+} frac{f(a+h) - f(a)}{h} = lim_{h o 0^+} frac{h(a+h) - h(a)}{h}$`.
Caution: You can only simply differentiate `g(x)` and `h(x)` and plug in `a` IF `g'(a)` and `h'(a)` are themselves defined using the standard differentiation rules. For instance, if `g(x) = |x-a|` and you are checking at `x=a`, you cannot just say `g'(a)` is `1` or `-1`. You MUST use the limit definition for the LHD/RHD at the point `a`. However, if `g(x)` and `h(x)` are polynomials or other standard differentiable functions, then `g'(a)` and `h'(a)` are valid LHD/RHD.

3. Equate LHD and RHD:
* If LHD = RHD, then `f(x)` is differentiable at `x=a`. The derivative `f'(a)` will be this common value.

#### Example 1: Differentiability of `f(x) = |x-2|` at `x=2`

Let `f(x) = |x-2|`. We want to check differentiability at `x=2`.
First, write `f(x)` as a piecewise function:
`f(x) = { -(x-2) for x < 2; (x-2) for x โ‰ฅ 2 }`

Step 1: Check Continuity at `x=2`
* `f(2) = (2-2) = 0`
* `$lim_{x o 2^-} f(x) = lim_{x o 2^-} -(x-2) = -(2-2) = 0$`
* `$lim_{x o 2^+} f(x) = lim_{x o 2^+} (x-2) = (2-2) = 0$`
Since LHL = RHL = `f(2)`, `f(x)` is continuous at `x=2`. So, we proceed.

Step 2: Check LHD and RHD at `x=2`
* LHD:
$$f'(2^-) = lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{-( (2+h)-2 ) - 0}{h}$$
$$= lim_{h o 0^-} frac{-h}{h} = lim_{h o 0^-} (-1) = -1$$
* RHD:
$$f'(2^+) = lim_{h o 0^+} frac{f(2+h) - f(2)}{h} = lim_{h o 0^+} frac{( (2+h)-2 ) - 0}{h}$$
$$= lim_{h o 0^+} frac{h}{h} = lim_{h o 0^+} (1) = 1$$

Step 3: Compare LHD and RHD
Here, LHD = -1 and RHD = 1. Since LHD โ‰  RHD, `f(x) = |x-2|` is not differentiable at `x=2`. This confirms our geometric understanding of sharp corners.

#### Example 2: Finding parameters for Differentiability

Find the values of `a` and `b` such that the function `f(x)` is differentiable at `x=1`:
`f(x) = { x^2 + 3x + a, for x โ‰ค 1; bx + 2, for x > 1 }`

Step 1: Continuity at `x=1`
For `f(x)` to be differentiable, it must first be continuous at `x=1`.
* `f(1) = 1^2 + 3(1) + a = 1 + 3 + a = 4 + a`
* `$lim_{x o 1^-} f(x) = lim_{x o 1^-} (x^2 + 3x + a) = 1^2 + 3(1) + a = 4 + a$`
* `$lim_{x o 1^+} f(x) = lim_{x o 1^+} (bx + 2) = b(1) + 2 = b + 2$`
For continuity, LHL = RHL = `f(1)`:
`4 + a = b + 2`
This gives our first equation: `a - b = -2` (Equation 1)

Step 2: Differentiability (LHD = RHD) at `x=1`
Now, we find the derivatives of the individual pieces:
* For `x < 1`, `f'(x) = d/dx (x^2 + 3x + a) = 2x + 3`
* For `x > 1`, `f'(x) = d/dx (bx + 2) = b`

Let's find the LHD and RHD at `x=1`:
* LHD at `x=1`: We use the derivative of the left piece and evaluate at `x=1`.
`$f'(1^-) = lim_{x o 1^-} (2x + 3) = 2(1) + 3 = 5$`
* RHD at `x=1`: We use the derivative of the right piece and evaluate at `x=1`.
`$f'(1^+) = lim_{x o 1^+} (b) = b$`

For differentiability, LHD = RHD:
`5 = b`
So, `b = 5` (Equation 2)

Step 3: Solve for `a` and `b`
Substitute `b=5` into Equation 1:
`a - 5 = -2`
`a = -2 + 5`
`a = 3`

Thus, for `f(x)` to be differentiable at `x=1`, we must have `a=3` and `b=5`.

### 8. JEE Advanced Focus and Key Takeaways

* JEE Advanced often tests differentiability on functions involving `|f(x)|`, `max(f(x), g(x))`, `min(f(x), g(x))`, or functions defined using `[x]` (greatest integer) or `{x}` (fractional part). These functions frequently introduce sharp corners or discontinuities.
* Modulus Functions: `f(x) = |g(x)|` is typically not differentiable at points where `g(x) = 0` and `g'(x) โ‰  0`. These points correspond to sharp corners. If `g(x) = 0` and `g'(x) = 0` (e.g., `f(x) = |x^2| = x^2`), then it can be differentiable.
* Greatest Integer Function: `[x]` is discontinuous at every integer, hence not differentiable at any integer.
* Fractional Part Function: `{x}` is discontinuous at every integer, hence not differentiable at any integer.
* Product/Quotient/Sum of Differentiable/Non-Differentiable Functions:
* Sum/Difference: If `f` is differentiable and `g` is not, `fยฑg` is generally not differentiable. If both are non-differentiable, `fยฑg` *could* be differentiable (e.g., `|x| + (-|x|) = 0`, which is differentiable).
* Product/Quotient: Similar complexities. For example, `f(x) = x|x|` is differentiable at `x=0`. `f(x) = x^2 |x|` is also differentiable. This is because the non-differentiability at `x=0` is 'cancelled out' by the `x` or `x^2` term making the LHD/RHD equal.
* Chain Rule in Differentiability: If `f(g(x))` is differentiable, `g(x)` must be continuous. If `g(x)` is differentiable at `a` and `f(u)` is differentiable at `u=g(a)`, then `f(g(x))` is differentiable at `a`. Be cautious when inner functions are non-differentiable.

Understanding differentiability is not just about computing derivatives; it's about understanding the very nature of a function's behavior โ€“ its smoothness, its ability to have a unique tangent, and its well-defined rate of change. Master these fundamental concepts, and you'll be well-equipped for the more complex problems awaiting you in JEE.
๐ŸŽฏ Shortcuts

Mnemonics & Shortcuts for Differentiability



Differentiability is a crucial yet often tricky concept in Calculus for JEE. Mastering it requires a clear understanding of its definition and its relationship with continuity. Here are some mnemonics and shortcuts to help you remember key aspects and quickly identify non-differentiable points, saving valuable time in exams.

1. Defining Differentiability: LHD = RHD


The core definition of a function `f(x)` being differentiable at `x=a` is that its Left Hand Derivative (LHD) equals its Right Hand Derivative (RHD) at that point.



  • Mnemonic: LHD RHD Match, Differentiability's Catch!

    • If LHD = RHD, the function is smooth at `x=a` and differentiable.

    • If LHD ≠ RHD, there's a sharp corner or a cusp, and the function is not differentiable.




  • Shortcut for Piecewise Functions (JEE Tip):
    For a piecewise function defined as `f(x) = g(x)` for `x < a` and `h(x)` for `x โ‰ฅ a` at a joint point `x=a`:

    1. First Check Continuity: Ensure `lim (x->a-) g(x) = lim (x->a+) h(x) = h(a)`. If not continuous, it's definitely not differentiable.

    2. Then Check Differentiability: Calculate `g'(a)` (LHD) and `h'(a)` (RHD). If `g'(a) = h'(a)`, then `f(x)` is differentiable at `x=a`. This often simplifies to directly differentiating each piece and checking the values at the joint point.





2. Differentiability vs. Continuity Relationship


This is a frequently tested concept, and students often confuse the implications.



  • Mnemonic: "Doctors are always Cheerful, but Cheerful people aren't always Doctors."

    • Differentiable &implies; Continuous (Doctors are always Cheerful).

    • Continuous ¬&implies; Differentiable (Cheerful people aren't always Doctors).


    This means if a function is differentiable at a point, it must be continuous there. However, a continuous function may or may not be differentiable (e.g., `f(x) = |x|` at `x=0` is continuous but not differentiable).



3. Identifying Points of Non-Differentiability


Knowing where differentiability typically fails can save a lot of time in exams.



  • Mnemonic for Failure Points: "C-C-D-V: Calculus Comes Difficult, Very! (at these points)"

    • Corners: Like `f(x) = |x|` at `x=0`. The slope changes abruptly. (LHD ≠ RHD)

    • Cusps: Like `f(x) = x^(2/3)` at `x=0`. The tangent becomes vertical, but the slope from left and right tends to ±∞. (LHD and RHD are infinite but different in nature)

    • Discontinuities: Jumps, holes, or vertical asymptotes. If a function is not continuous, it cannot be differentiable.

    • Vertical Tangents: Like `f(x) = x^(1/3)` at `x=0`. The tangent line is vertical, meaning the derivative is infinite. (LHD and RHD are both ±∞)




  • Shortcut for Absolute Value Functions `|f(x)|` (JEE Tip):
    A function of the form `F(x) = |f(x)|` is typically non-differentiable at points `x` where `f(x) = 0` AND `f'(x) ≠ 0`. These are the "sharp corners" created by the absolute value.

    • If `f(x) = 0` and `f'(x) = 0` at a point (e.g., `f(x) = x^2` at `x=0`), then `|f(x)|` *might* be differentiable (e.g., `|x^2| = x^2` is differentiable at `x=0`). In such cases, a direct LHD/RHD check is needed.

    • Mnemonic: "Absolute Value's Zero `f` needs a Non-Zero `f'` to be a problem."




  • Specific Functions & Their Non-Differentiable Points:

    • Greatest Integer Function `[x]` (Floor function): Non-differentiable at all integer points `(..., -2, -1, 0, 1, 2, ...)` due to jump discontinuities.
      • Mnemonic: Think of `[x]` as a STAIRCASE; you can't draw a smooth tangent on a step.


    • Signum Function `sgn(x)`: Non-differentiable at `x=0` due to a jump discontinuity.
      • Mnemonic: `sgn(x)` makes a JUMP at zero.


    • Fractional Part Function `{x}`: Non-differentiable at all integer points due to jump discontinuities.





By internalizing these mnemonics and shortcuts, you can approach differentiability problems with greater confidence and efficiency in your exams. Focus on the visual interpretations of these concepts, as they often provide the quickest path to a solution.
๐Ÿ’ก Quick Tips

Quick Tips for Differentiability



Differentiability is a core concept in Calculus, forming the foundation for many advanced topics. For both JEE Main and Board exams, a clear understanding and efficient problem-solving strategy are crucial. These quick tips will help you navigate common problems and avoid pitfalls.



  • Understand the Core Definition: A function `f(x)` is differentiable at `x = a` if its derivative `f'(a)` exists. This means the Left Hand Derivative (LHD) must be equal to the Right Hand Derivative (RHD) and both must be finite.


    `LHD = lim (hโ†’0โป) [f(a+h) - f(a)] / h`


    `RHD = lim (hโ†’0โบ) [f(a+h) - f(a)] / h`


  • Visual Check (JEE & CBSE):

    • If a function's graph has a sharp corner or a cusp at a point, it is not differentiable at that point. (e.g., `|x|` at `x=0`).

    • If a function is discontinuous (has a break or jump) at a point, it is not differentiable at that point.

    • If a function has a vertical tangent at a point, it is not differentiable there (derivative is infinite).




  • Continuity is a Prerequisite: A function must be continuous at a point to be differentiable at that point. If a function is discontinuous at `x=a`, it is immediately not differentiable at `x=a`.


    Important: Differentiability implies continuity, but continuity does NOT imply differentiability. This is a frequently tested concept in JEE.


  • Handling Modulus Functions (JEE Focus):

    • Functions involving `|g(x)|` are often not differentiable where `g(x) = 0`. For example, `f(x) = |x-c|` is not differentiable at `x=c`.

    • To check differentiability, remove the modulus sign by considering cases `g(x) โ‰ฅ 0` and `g(x) < 0`, then apply LHD/RHD at `g(x)=0`.




  • Piecewise Defined Functions (JEE & CBSE):

    • Always check differentiability at the junction points (where the function definition changes).

    • First, ensure the function is continuous at the junction point. If not, it's not differentiable.

    • Then, calculate LHD and RHD using the limit definition or by differentiating the respective pieces and evaluating the limits.




  • Algebra of Differentiable Functions:

    • If `f(x)` and `g(x)` are differentiable at `x=a`, then `fยฑg`, `fยทg`, and `f/g` (if `g(a)โ‰ 0`) are also differentiable at `x=a`.

    • The composition `f(g(x))` is differentiable if `f` is differentiable at `g(a)` and `g` is differentiable at `a`.




  • JEE Specific Strategy for Difficult Functions:
    For functions involving `max(f(x), g(x))`, `min(f(x), g(x))`, `[x]` (Greatest Integer Function), or `{x}` (Fractional Part Function), direct application of LHD and RHD definitions is usually the safest approach at potential non-differentiable points. These functions are typically non-differentiable at points where their definition changes or at integer points for `[x]` and `{x}`.


  • Derivative of Inverse Function: If `y = f(x)` and `x = g(y)` are inverse functions, then `g'(y) = 1 / f'(x)`, provided `f'(x) โ‰  0`. Remember this relation when solving problems involving inverse trigonometric functions.



Mastering differentiability requires practice, especially with piecewise and modulus functions. Focus on understanding the conceptual meaning behind the formulas.
๐Ÿง  Intuitive Understanding

Intuitive Understanding of Differentiability


At its core, differentiability is a measure of how "smooth" a function's graph is at a particular point. Intuitively, a function is differentiable at a point if its graph has a well-defined, unique, and non-vertical tangent line at that specific point. This tangent line represents the instantaneous rate of change or the slope of the curve precisely at that point.



What Differentiability Means Graphically:




  • Smoothness: A differentiable function "flows" smoothly without any abrupt changes in direction. Imagine tracing the curve with a pen โ€“ if you can draw it without lifting the pen and without making any sharp turns or corners, it's likely differentiable.


  • Unique Tangent: At a point where a function is differentiable, you can draw one and only one non-vertical straight line that just "touches" the curve at that point without cutting through it (locally). The slope of this unique tangent is the derivative.



When a Function is NOT Differentiable (Intuitive Visuals):


A function fails to be differentiable at a point if:




  • Sharp Corner or Cusp: If the graph has a "pointy" corner (like the function $f(x) = |x|$ at $x=0$) or a cusp (like $f(x) = x^{2/3}$ at $x=0$), you cannot draw a unique tangent line. As you approach the corner from the left, the slope might be one value, and from the right, it's a different value. There's no single, clear tangent.


    Example: For $f(x)=|x|$ at $x=0$, the left-hand slope is $-1$, and the right-hand slope is $+1$. Since these are not equal, $f(x)=|x|$ is not differentiable at $x=0$.




  • Discontinuity (Break in the Graph): If the function is not continuous at a point (i.e., there's a jump, hole, or vertical asymptote), it cannot be differentiable at that point. You cannot draw a tangent line on a broken curve.


    Key Takeaway for CBSE & JEE: Differentiability implies continuity. If a function is differentiable at a point, it must be continuous at that point. However, the converse is not true; continuity does not guarantee differentiability (as seen with sharp corners).




  • Vertical Tangent: If the tangent line at a point is vertical (i.e., its slope is infinite or undefined, such as $f(x) = x^{1/3}$ at $x=0$), the function is not considered differentiable at that point. The derivative must be a finite real number.





The Role of Left-Hand and Right-Hand Derivatives:


Intuitively, for a function to be differentiable at a point, the "slope" as you approach that point from the left must be the same as the "slope" as you approach it from the right. These are called the Left-Hand Derivative (LHD) and Right-Hand Derivative (RHD), respectively. If LHD = RHD and both are finite, then the function is differentiable at that point.



JEE Main Perspective:


For JEE Main, a strong intuitive understanding, especially graphical interpretation, is vital. You'll often encounter problems where you need to identify points of non-differentiability by simply looking at the graph or analyzing the function for corners, cusps, or discontinuities. While formal definitions are important, the visual insight speeds up problem-solving significantly and aids in quickly checking options.

๐ŸŒ Real World Applications

While the IIT JEE and board exams primarily focus on the theoretical understanding and computational aspects of differentiability, recognizing its real-world relevance can deepen your intuition and appreciation for the concept. Differentiability, at its core, describes the smoothness of a function and the existence of a well-defined instantaneous rate of change at every point. This concept is crucial for modeling and analyzing systems where continuous and smooth changes are observed.



Here are some key real-world applications of differentiability:




  • Physics and Engineering:

    • Motion Analysis: If a particle's position is given by a function $s(t)$, its instantaneous velocity $v(t)$ is $s'(t)$ (the first derivative), and its instantaneous acceleration $a(t)$ is $v'(t) = s''(t)$ (the second derivative). Differentiability ensures smooth, physically realistic motion without instantaneous jumps in velocity.

    • Optimization: Engineers use differentiability to optimize designs, such as maximizing the efficiency of an engine, minimizing material usage in construction, or finding the most stable configuration of a structure. This involves finding critical points where the derivative is zero.

    • Control Systems: Designing systems that respond smoothly to inputs (e.g., cruise control in a car, robotic arm movements) relies on differentiable functions to describe the system's dynamics.



  • Economics and Business:

    • Marginal Analysis: In economics, concepts like marginal cost, marginal revenue, and marginal profit are derivatives of total cost, total revenue, and total profit functions, respectively. They represent the rate of change of these quantities with respect to an additional unit produced or sold. Businesses use this to make production and pricing decisions.

    • Elasticity: Measures like price elasticity of demand involve derivatives to quantify how responsive one variable is to changes in another.



  • Biology and Medicine:

    • Population Growth Models: Differentiable functions are used to model the growth or decay of populations (e.g., bacteria, animal species) or the spread of diseases. The derivative gives the instantaneous growth rate.

    • Drug Concentration: Modeling how drug concentration in the bloodstream changes over time after administration, where the rate of change is described by a derivative.



  • Computer Graphics and Animation:

    • Smooth Transitions: To create realistic and smooth animations, camera movements, or object transformations, the paths and trajectories are often defined by differentiable functions. This prevents jerky or sudden changes, making the animation appear fluid.

    • Surface Modeling: Differentiability is used in rendering smooth surfaces and curves, ensuring continuity and seamless transitions in 3D models.



  • Data Science and Machine Learning:

    • Gradient Descent: This is a fundamental optimization algorithm used to train many machine learning models. It relies heavily on calculating the gradient (a vector of partial derivatives) of a cost function to find its minimum, thereby optimizing model parameters. The ability to differentiate the cost function is crucial for this process.





JEE Relevance: While specific real-world problems involving complex modeling might not appear directly in JEE, understanding these applications reinforces the core concept that differentiability signifies a "well-behaved" function whose instantaneous rate of change can be precisely determined. This intuition is invaluable when interpreting results from derivative calculations in problem-solving.

๐Ÿ”„ Common Analogies

Understanding differentiability can sometimes be abstract. Analogies help build intuition by relating complex mathematical concepts to everyday experiences. For JEE and Board exams, while formal definitions are paramount, these analogies can solidify your grasp of the underlying ideas.



Common Analogies for Differentiability





  1. Driving a Car on a Road:

    • Imagine driving a car along a road which represents the graph of a function.

    • If the road is smooth and continuous, your steering wheel makes gradual, continuous adjustments. At any given instant, there's a unique, definite direction you're heading (the slope of the tangent). This is analogous to a function being differentiable.

    • If the road has a sharp corner or a cusp (like a V-shape turn), you can't smoothly navigate it. You have to make an abrupt, instantaneous change in your steering, or it's ambiguous which exact direction you're facing *at that precise corner*. This point represents a location where the function is not differentiable.

    • If there's a break in the road or a cliff (a discontinuity), you cannot drive across it. Similarly, a function cannot be differentiable at a point where it's discontinuous.

    • This analogy highlights that differentiability implies "smoothness" and "no sharp turns."




  2. Zooming In (Local Linearity):

    • Consider a very powerful magnifying glass. When you look at a tiny segment of a differentiable curve through this glass, as you zoom in more and more, that segment will appear increasingly like a straight line. The slope of this "apparent" straight line is the derivative at that point.

    • This means that locally, around a differentiable point, the curve behaves very much like its tangent line.

    • However, if you zoom in on a sharp corner or a cusp, no matter how much you magnify it, it will *never* look like a straight line; the "pointiness" will always be visible. This indicates non-differentiability.

    • This analogy is crucial for understanding the concept of a tangent as the "best linear approximation" of the curve at a point.




  3. Rolling a Ball on a Surface:

    • Imagine rolling a small ball along the graph of a function.

    • If the curve is smooth, the ball will roll seamlessly, and at any point, its instantaneous direction of travel (tangent) is uniquely defined. This signifies differentiability.

    • If the curve has a sharp peak or trough (cusp), the ball would momentarily hesitate or change direction abruptly, lacking a single, clear direction of motion *at that exact point*.

    • If there's a gap or a vertical jump (discontinuity), the ball cannot traverse that point smoothly or at all.





JEE & CBSE Callout: While these analogies build strong intuition, remember that formal definitions (left-hand derivative, right-hand derivative, limit of the difference quotient) are what you'll use for solving problems and proving differentiability in exams. These analogies help you visualize and anticipate the nature of differentiability at various points on a graph.

๐Ÿ“‹ Prerequisites

Welcome to the 'Prerequisites' section for Differentiability! Before diving into the intricacies of derivatives, it's crucial to have a strong foundation in a few core concepts. Differentiability builds directly upon these ideas, and a shaky understanding here can lead to significant difficulties later on. This section outlines the essential topics you must master to confidently tackle Differentiability.



Essential Prerequisite Concepts:





  1. Functions and their Properties:

    • Understanding Functions: A clear grasp of what a function is, its domain, range, and co-domain.

    • Types of Functions: Familiarity with common function types such as polynomial, trigonometric, exponential, logarithmic, modulus functions (|x|), greatest integer function ([x]), and signum function (sgn(x)). Understanding their graphs and behaviors is particularly important, as these are frequently used in differentiability problems to identify points of non-differentiability (e.g., sharp corners, jumps).




  2. Limits:

    • Definition and Existence of Limits: This is arguably the most critical prerequisite. Differentiability is fundamentally defined using limits. You must understand that a limit exists if and only if the Left Hand Limit (LHL) equals the Right Hand Limit (RHL) at a given point.

    • Evaluation Techniques: Proficiency in evaluating limits using various methods like factorization, rationalization, using standard limits (e.g., $lim_{x o 0} frac{sin x}{x} = 1$), and L'Hopital's Rule (though L'Hopital's is often taught after differentiation, basic limit evaluation techniques are a must).

    • Geometric Interpretation: An intuitive understanding of limits as values a function approaches.

    • JEE Focus: JEE heavily tests limits. A strong command over limit evaluation, especially for indeterminate forms, is non-negotiable for success in differentiability problems.




  3. Continuity:

    • Definition of Continuity: A function $f(x)$ is continuous at a point $x=a$ if $lim_{x o a^-} f(x) = lim_{x o a^+} f(x) = f(a)$.

    • Types of Discontinuities: Knowledge of removable and non-removable discontinuities (jump, infinite, oscillatory). Understanding these helps in quickly identifying points where a function cannot be differentiable.

    • Relationship with Differentiability: A fundamental theorem states that if a function is differentiable at a point, it must be continuous at that point. However, the converse is not true (a continuous function may not be differentiable). This relationship is frequently tested in both CBSE and JEE.

    • JEE Focus: Questions often combine continuity and differentiability, asking students to determine if a function is continuous and/or differentiable at certain points, especially for piecewise defined functions.




  4. Basic Algebra and Trigonometry:

    • Algebraic Manipulation: Proficiency in simplifying algebraic expressions, factoring, and solving equations. These skills are constantly used when calculating and analyzing derivatives.

    • Trigonometric Identities: A good recall of fundamental trigonometric identities is essential, as derivatives of trigonometric functions often require simplification using these identities.





Before moving forward, ensure you can confidently solve problems related to these prerequisite topics. Revisit your notes and practice questions on limits and continuity to build a solid foundation. This investment will pay off significantly as you delve deeper into Differentiability.

โš ๏ธ Common Exam Traps

Common Exam Traps in Differentiability



Understanding differentiability is crucial for calculus, and examiners often set traps to test a student's conceptual clarity. Being aware of these common pitfalls can significantly improve your performance.



  • Confusing Continuity with Differentiability:

    This is arguably the most frequent trap. Many students mistakenly assume that if a function is continuous at a point, it must be differentiable there. This is incorrect.



    • The Truth: Differentiability implies continuity. If a function is differentiable at a point, it *must* be continuous at that point.

    • The Trap: The converse is not true. A function can be continuous but not differentiable at a point. The classic example is $f(x) = |x|$ at $x=0$. It is continuous at $x=0$ but has a sharp corner, making it non-differentiable.

    • How to Avoid: Always remember that a "sharp turn" or "cusp" signifies non-differentiability, even if the function is continuous.




  • Incorrect Handling of Piecewise Functions:

    For piecewise functions, students often forget to check differentiability at the "junction points" (where the function definition changes). They might only check continuity.



    • The Trap: To check differentiability at a junction point, say $x=a$, you must verify two conditions:

      1. The function must be continuous at $x=a$. (i.e., $LHL = RHL = f(a)$)

      2. The Left Hand Derivative (LHD) must be equal to the Right Hand Derivative (RHD) at $x=a$.


      Many students only perform step 1 or make an error in step 2 by directly differentiating each piece without using the limit definition for LHD/RHD at the junction.

    • How to Avoid (JEE Specific): Always use the definition of derivative, i.e., $f'(a) = lim_{h o 0} frac{f(a+h) - f(a)}{h}$ to find LHD and RHD at the junction points. For instance, LHD = $lim_{h o 0^-} frac{f(a+h) - f(a)}{h}$ and RHD = $lim_{h o 0^+} frac{f(a+h) - f(a)}{h}$.




  • Misinterpreting Absolute Value Functions:

    Functions involving absolute values, like $f(x) = |g(x)|$, are frequently used to test differentiability.



    • The Trap: Students often forget that $f(x) = |g(x)|$ is generally not differentiable at points where $g(x) = 0$ and $g'(x)
      eq 0$. This creates a sharp corner (e.g., $|x-a|$ at $x=a$).

    • How to Avoid: Convert the absolute value function into a piecewise function by considering the cases where the argument inside the absolute value is positive, negative, or zero. Then apply the rules for piecewise functions. For instance, for $f(x) = |x-2|(x+1)$, define it as:

      • $-(x-2)(x+1)$ for $x < 2$

      • $(x-2)(x+1)$ for $x ge 2$


      Then check differentiability at $x=2$.




  • Domain Considerations for Differentiability:

    Differentiability can only be discussed for points within the domain of the function, and specifically at interior points of the domain.



    • The Trap: For functions like $f(x) = sqrt{x}$, students might incorrectly state it's differentiable at $x=0$. However, the Left Hand Derivative at $x=0$ is not defined because the function is not defined for $x<0$.

    • How to Avoid: Always check the domain of the function first. Differentiability at endpoints of a closed interval requires specific definitions (e.g., only RHD at the left endpoint, only LHD at the right endpoint).




  • Algebraic Errors in LHD/RHD Limits:

    Even with correct conceptual understanding, evaluating the limits for LHD and RHD can be algebraically intensive, leading to mistakes.



    • The Trap: Complex expressions, rationalization, or application of standard limits/L'Hopital's rule can introduce computational errors.

    • How to Avoid: Practice limit evaluation extensively. Be meticulous with algebraic manipulations. Double-check your calculations, especially when dealing with square roots or fractional powers.





💪 JEE Tip: Mastering differentiability for piecewise and absolute value functions is a recurring theme. Focus on these areas for exam success!

โญ Key Takeaways

Key Takeaways: Differentiability



Differentiability is a cornerstone concept in calculus, central to understanding the behavior of functions. It extends the idea of continuity by introducing the concept of a "smooth" curve. Mastering differentiability is crucial for both JEE Main and board examinations.

Here are the essential takeaways:



  • Definition of Differentiability at a Point:
    A function (f(x)) is said to be differentiable at a point (x=a) if the derivative (f'(a)) exists. This means the limit
    [ lim_{h o 0} frac{f(a+h) - f(a)}{h} ]
    exists and is finite.


  • Left Hand Derivative (LHD) and Right Hand Derivative (RHD):
    For the derivative to exist at (x=a), both the Left Hand Derivative (LHD) and the Right Hand Derivative (RHD) must exist, be finite, and be equal.


    • LHD at (x=a): (f'(a^-) = lim_{h o 0^-} frac{f(a+h) - f(a)}{h} = lim_{h o 0} frac{f(a-h) - f(a)}{-h})


    • RHD at (x=a): (f'(a^+) = lim_{h o 0^+} frac{f(a+h) - f(a)}{h})


    • Condition for Differentiability: A function (f(x)) is differentiable at (x=a) if and only if (f'(a^-) = f'(a^+) = ext{a finite value}). This explicit check is frequently required in JEE Main problems, especially for piecewise functions.




  • Geometric Interpretation:
    A function is differentiable at a point if and only if there exists a unique (non-vertical) tangent line to the graph of the function at that point.


    • If LHD (
      eq ) RHD, the curve has a "corner" or "cusp" at that point (e.g., (f(x) = |x|) at (x=0)).


    • If the tangent is vertical, the derivative is infinite, so the function is not differentiable (e.g., (f(x) = x^{1/3}) at (x=0)).




  • Relationship Between Differentiability and Continuity:
    This is a critical concept for both CBSE Board and JEE Main exams:


    • Differentiability implies Continuity: If a function is differentiable at a point, it is necessarily continuous at that point. This is a fundamental theorem.


    • Continuity does NOT imply Differentiability: A function can be continuous at a point but not differentiable at that point. The classic example is (f(x) = |x|) at (x=0), which is continuous but has a sharp corner, hence not differentiable.




  • Points Where Differentiability Fails:
    A function cannot be differentiable at points where its graph has:

    • Discontinuities: If a function is not continuous at a point, it cannot be differentiable there.

    • Sharp corners or kinks: Where the tangent line is not unique (e.g., (|x|), (cos|x|)).

    • Cusps: Where the tangent line becomes vertical from both sides (e.g., (x^{2/3})).

    • Vertical Tangents: Where the derivative is infinite (e.g., (x^{1/3})).




  • Practical Approach for Problems (JEE Focus):
    For functions, especially piecewise defined ones, to check differentiability at the "junction" point (x=a):

    1. First, check for continuity at (x=a). If it's not continuous, it's not differentiable.

    2. If it's continuous, then calculate LHD and RHD at (x=a).

    3. If LHD = RHD = finite value, then the function is differentiable at (x=a).





Understanding these key takeaways will provide a robust foundation for solving a wide range of problems related to differentiability in your exams. Practice is key to internalizing these concepts!

๐Ÿงฉ Problem Solving Approach

๐Ÿš€ Problem-Solving Approach: Differentiability


Mastering differentiability problems requires a systematic approach, especially at critical points or for piecewise functions. This section outlines a step-by-step strategy to confidently tackle such questions in both Board Exams and JEE.



Step 1: Initial Check - Continuity (Crucial for JEE)


Before checking for differentiability, always verify if the function is continuous at the point in question. Differentiability implies continuity. If a function is not continuous at a point, it cannot be differentiable there. This is a common first filter for JEE problems.



  • A function $f(x)$ is continuous at $x=a$ if $lim_{x o a^-} f(x) = lim_{x o a^+} f(x) = f(a)$.

  • If this condition fails, the function is immediately non-differentiable at $x=a$.



Step 2: Identify Critical Points


Differentiability issues typically arise at:



  • Points where the function's definition changes (for piecewise functions).

  • Points where expressions inside modulus functions become zero (e.g., $|f(x)|$ is problematic where $f(x)=0$).

  • Points where functions like Greatest Integer Function ([x]) or Fractional Part Function ({x}) are involved (they are generally non-differentiable at integer points).

  • Points where the derivative of an elementary function might be undefined (e.g., $f'(x)$ being undefined due to division by zero or negative under a square root).



Step 3: Apply the First Principle of Differentiation (LHD & RHD)


This is the most robust method for determining differentiability, especially at critical points or for piecewise functions.



  • Left Hand Derivative (LHD) at $x=a$: $f'(a^-) = lim_{h o 0^-} frac{f(a+h) - f(a)}{h}$

  • Right Hand Derivative (RHD) at $x=a$: $f'(a^+) = lim_{h o 0^+} frac{f(a+h) - f(a)}{h}$

  • For $f(x)$ to be differentiable at $x=a$, both LHD and RHD must exist, be finite, and be equal ($f'(a^-) = f'(a^+)$).



Step 4: Using Direct Differentiation (where applicable)


For functions defined by a single formula (not piecewise), or within intervals where they behave as elementary functions:



  • Differentiate $f(x)$ to find $f'(x)$ using standard differentiation rules.

  • Check if $f'(x)$ is defined and finite at the point in question. If $f'(x)$ exists at $x=a$ and the function is continuous at $x=a$, then it is differentiable at $x=a$.

  • JEE Tip: Often, for piecewise functions, you can find the derivatives of each piece and then check if the left and right derivatives match at the junction point. However, this shortcut implicitly assumes continuity, so an explicit continuity check (Step 1) is still safer.



Example: Checking Differentiability of a Piecewise Function


Let's check the differentiability of $f(x) = egin{cases} x^2 & ext{if } x ge 0 \ x & ext{if } x < 0 end{cases}$ at $x=0$.



  1. Continuity at $x=0$:

    • $f(0) = 0^2 = 0$

    • $lim_{x o 0^-} f(x) = lim_{x o 0^-} x = 0$

    • $lim_{x o 0^+} f(x) = lim_{x o 0^+} x^2 = 0$


    Since $f(0) = lim_{x o 0^-} f(x) = lim_{x o 0^+} f(x) = 0$, $f(x)$ is continuous at $x=0$. Proceed to differentiability.

  2. LHD at $x=0$:
    $f'(0^-) = lim_{h o 0^-} frac{f(0+h) - f(0)}{h} = lim_{h o 0^-} frac{(0+h) - 0}{h} = lim_{h o 0^-} frac{h}{h} = 1$.

  3. RHD at $x=0$:
    $f'(0^+) = lim_{h o 0^+} frac{f(0+h) - f(0)}{h} = lim_{h o 0^+} frac{(0+h)^2 - 0}{h} = lim_{h o 0^+} frac{h^2}{h} = lim_{h o 0^+} h = 0$.


Since LHD ($1$) $
eq$ RHD ($0$), the function $f(x)$ is not differentiable at $x=0$, even though it is continuous there. This visually corresponds to a sharp corner at $x=0$.



Key Takeaways for JEE:



  • Always prioritize checking continuity first. It's a faster way to rule out differentiability.

  • For functions involving $|g(x)|$, test differentiability at points where $g(x)=0$. The derivative will often change sign, leading to non-differentiability (a cusp or sharp corner).

  • Remember that the sum, difference, product, and composition of differentiable functions are differentiable. Problems often involve checking differentiability where this property might break down due to non-differentiable components (e.g., $f(x) = x|x|$ is differentiable at $x=0$ despite $|x|$ not being).


๐Ÿ“ CBSE Focus Areas

For students preparing for the CBSE Board Examinations, understanding Differentiability is crucial. While the core mathematical concepts align with JEE, the CBSE curriculum often focuses on specific types of problems, step-by-step verification, and the application of differentiation rules.



Core Concepts for CBSE


In CBSE, differentiability at a point x=a for a function f(x) is fundamentally checked by evaluating the existence and equality of its Left-Hand Derivative (LHD) and Right-Hand Derivative (RHD).



  • Left-Hand Derivative (LHD): fโ€ฒ(aโˆ’)=limhโ†’0+f(a)โˆ’f(aโˆ’h)h or limhโ†’0โˆ’f(a+h)โˆ’f(a)h

  • Right-Hand Derivative (RHD): fโ€ฒ(a+)=limhโ†’0+f(a+h)โˆ’f(a)h


A function f(x) is differentiable at x=a if and only if LHD = RHD = finite value.



Key Conditions and Implications



  1. Continuity is Necessary: If a function is differentiable at a point, it must be continuous at that point. However, the converse is not true; a continuous function may not be differentiable (e.g., f(x)=|x| at x=0). CBSE often tests this relationship.

  2. Geometric Interpretation: Differentiability implies a smooth curve without sharp corners (cusps) or vertical tangents at that point.



Common CBSE Problem Types


CBSE questions on differentiability often fall into these categories:




  • Checking Differentiability of Piecewise Functions:

    Given a function defined in parts (e.g., f(x)={x2xโ‰ค1xx>1), you will be asked to check its differentiability at the critical point(s) where the definition changes. This typically involves calculating LHD and RHD explicitly using the limit definition.




  • Finding Constants for Differentiability:

    A common problem involves finding values of constants (e.g., a,b) such that a piecewise function is differentiable at a given point. This requires setting up two equations:



    1. One for continuity at the point (LHL = RHL = f(point)).

    2. One for differentiability at the point (LHD = RHD).




  • Differentiability of Functions involving Absolute Values:

    Functions like f(x)=|xโˆ’a| are not differentiable at x=a because of a sharp corner. CBSE expects you to demonstrate this using LHD and RHD, or by redefining the function without the absolute value sign.





CBSE vs. JEE Perspective























Aspect CBSE Board Exams JEE Main/Advanced
Focus Emphasis on procedural steps, LHD/RHD calculations, application of standard rules, and clear presentation. Tests deeper conceptual understanding, non-trivial functions (e.g., involving Greatest Integer Function, Fractional Part Function), graphical analysis, and implications of non-differentiability.
Question Complexity Generally straightforward piecewise functions, absolute value functions, and finding constants. More abstract functions, composite functions, and questions often combine differentiability with other concepts like limits, continuity, and properties of functions.


Exam Tip: For CBSE, always show all steps clearly when calculating LHD and RHD using the limit definition. Do not jump to conclusions based on mental calculations. Practice a variety of piecewise functions to master finding unknown constants.

๐ŸŽ“ JEE Focus Areas

๐ŸŽฏ JEE Focus Areas: Differentiability


Understanding differentiability is critical for JEE Main, forming a bridge between continuity and the core of calculus. JEE questions often test your ability to determine differentiability, especially for piecewise functions or functions involving absolute values, and to find unknown parameters that make a function differentiable.



Key Concepts & JEE Approach



  • Definition of Differentiability: A function f ( x ) is differentiable at x = a if the derivative f โ€ฒ ( a ) exists and is finite. This means the Left Hand Derivative (LHD) and Right Hand Derivative (RHD) at x = a must be equal and finite.

    • LHD at x = a : f โ€ฒ ( a โˆ’ ) = lim h โ†’ 0 + f ( a โˆ’ h ) โˆ’ f ( a ) โˆ’ h

    • RHD at x = a : f โ€ฒ ( a + ) = lim h โ†’ 0 + f ( a + h ) โˆ’ f ( a ) h



  • Continuity is a Prerequisite: If a function is differentiable at a point, it must be continuous at that point. However, the converse is not true (a continuous function may not be differentiable, e.g., f ( x ) = | x | at x = 0 ).

    • JEE Tip: Before checking for differentiability, always verify continuity. If a function is not continuous at a point, it cannot be differentiable there.



  • Geometric Interpretation: Differentiability at a point implies the existence of a unique tangent line at that point. Non-differentiability occurs at sharp corners (cusps), vertical tangents, or points of discontinuity.



Common JEE Problem Types



  1. Piecewise Defined Functions: These are the most frequent. To check differentiability at the "junction" point:

    • First, ensure the function is continuous at that point (LHL = RHL = function value). This often gives one equation involving parameters.

    • Then, calculate LHD and RHD using the limit definition or by differentiating each piece and evaluating at the junction point (if the derivatives exist). Equating LHD and RHD provides another equation.

    • Solve the system of equations for the unknown parameters.



  2. Functions Involving Modulus: | g ( x ) | is generally non-differentiable where g ( x ) = 0 , unless g โ€ฒ ( x ) = 0 at that point. Rewrite the function as a piecewise function to analyze.

  3. Functions Involving Greatest Integer Function (GIF) and Fractional Part Function (FPF): These functions are typically non-differentiable at integer points due to their inherent discontinuities or sharp jumps.

  4. Differentiability of Composite Functions: For functions like f ( g ( x ) ) , differentiability depends on both f and g . Special attention is needed when inner or outer functions are not differentiable at certain points.



Illustrative Example (JEE Type)


Problem: Find the values of a and b such that the function f ( x ) = x 2 + 3 x + a , x โ‰ค 1 b x + 2 , x > 1 is differentiable at x = 1 .


Solution:



  1. Continuity at x = 1 :

    lim x โ†’ 1 โˆ’ f ( x ) = 1 2 + 3 ( 1 ) + a = 4 + a

    lim x โ†’ 1 + f ( x ) = b ( 1 ) + 2 = b + 2

    f ( 1 ) = 1 2 + 3 ( 1 ) + a = 4 + a

    For continuity, 4 + a = b + 2 โŸน a โˆ’ b = โˆ’ 2 (Eq. 1)

  2. Differentiability at x = 1 :

    First, find the derivatives of each piece:

    For x < 1 , f โ€ฒ ( x ) = 2 x + 3 . So, LHD at x = 1 is 2 ( 1 ) + 3 = 5 .

    For x > 1 , f โ€ฒ ( x ) = b . So, RHD at x = 1 is b .

    For differentiability, LHD = RHD โŸน 5 = b . (Eq. 2)

  3. Solving for a and b :

    Substitute b = 5 into Eq. 1: a โˆ’ 5 = โˆ’ 2 โŸน a = 3 .

    Thus, a = 3 and b = 5 .



JEE vs. CBSE Callout


While CBSE board exams might primarily focus on the application of the derivative definition for standard functions, JEE Main extensively uses the concepts of continuity and differentiability together. Parameters in piecewise functions or functions with absolute values are a classic JEE staple. Always remember the "differentiable implies continuous" rule as a powerful first check.



Mastering these areas will significantly boost your performance in JEE questions on Differentiability!


๐ŸŒ Overview
A function is differentiable at x=a if the derivative f'(a) exists (finite and unique). Differentiability โ‡’ continuity, but not conversely. Non-differentiability can arise at corners, cusps, vertical tangents, or discontinuities; left/right derivatives must match.
๐Ÿ“š Fundamentals
โ€ข Differentiable โ‡’ continuous; converse false.
โ€ข Left/right derivatives must be equal and finite.
โ€ข Common tests at junctions of piecewise functions.
๐Ÿ”ฌ Deep Dive
Functions continuous everywhere but nowhere differentiable (qualitative); Lipschitz vs Hรถlder continuity; role in optimization.
๐ŸŽฏ Shortcuts
โ€œDโ‡’Cโ€ โ€” Differentiable implies Continuous (but not vice versa).
๐Ÿ’ก Quick Tips
โ€ข Absolute value: non-differentiable at 0.
โ€ข For |x|^n, differentiable at 0 for n>1 but not for n=1.
โ€ข Check vertical tangents where f'(a) may blow up.
๐Ÿง  Intuitive Understanding
Differentiability means a well-defined tangent slope at the point. Sharp turns or vertical tangents break this smooth, linear approximation.
๐ŸŒ Real World Applications
โ€ข Smooth motion implies well-defined instantaneous velocity.
โ€ข Optimization methods rely on derivatives near optima.
โ€ข Curve fitting with smoothness constraints.
๐Ÿ”„ Common Analogies
โ€ข โ€œZoom-in straightnessโ€: if you zoom sufficiently, the curve looks like a straight line around the point (has a unique slope).
๐Ÿ“‹ Prerequisites
Limits, continuity, difference quotient definition, and understanding of left/right limits.
โš ๏ธ Common Exam Traps
โ€ข Assuming continuity โ‡’ differentiability.
โ€ข Ignoring one-sided derivatives.
โ€ข Treating infinite slope as differentiable with finite derivative.
โญ Key Takeaways
โ€ข Derivative is a local linear model.
โ€ข Corners/cusps/vertical tangents break differentiability.
โ€ข Always check one-sided derivatives for piecewise definitions.
๐Ÿงฉ Problem Solving Approach
1) Compute LHD and RHD via limits.
2) Compare values to confirm equality/finitemess.
3) For parameters, enforce matching derivatives along with continuity.
4) Classify the failure mode if not differentiable.
๐Ÿ“ CBSE Focus Areas
Definition via limits; examples of differentiable vs non-differentiable points; relation to continuity.
๐ŸŽ“ JEE Focus Areas
Parameter problems enforcing differentiability at junctions; absolute value/floor effects; vertical tangent detection.

No CBSE problems available yet.

No JEE problems available yet.

No videos available yet.

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๐Ÿ“Important Formulas (4)

Derivative Definition (First Principle)
frac{dy}{dx} = f'(x) = lim_{h o 0} frac{f(x+h) - f(x)}{h}
Text: The derivative of a function f(x) is defined as the limit of the difference quotient as the change in x (denoted by h) approaches zero.
This is the fundamental limit definition used to find the instantaneous rate of change or the slope of the tangent to the curve at a point $x$. If this limit exists and is finite, the function is differentiable at $x$. <strong>(JEE/CBSE Significance: Essential for proofs and calculating derivatives from scratch.)</strong>
Variables: To calculate the derivative of a function $f(x)$ based purely on its definition, or when deriving standard differentiation rules.
Left Hand Derivative (LHD)
LHD = f'(a^-) = lim_{h o 0^+} frac{f(a) - f(a-h)}{h} quad ext{or} quad lim_{h o 0^-} frac{f(a+h) - f(a)}{h}
Text: The LHD is the limit of the slope approached from values slightly less than the point 'a'.
LHD measures the slope of the tangent immediately to the left of the point $x=a$. It is crucial when dealing with functions defined piecewise or involving modulus signs (like $f(x) = |x|$).
Variables: To check differentiability at a specific point $x=a$ for a piecewise function. Must be calculated along with RHD.
Right Hand Derivative (RHD)
RHD = f'(a^+) = lim_{h o 0^+} frac{f(a+h) - f(a)}{h}
Text: The RHD is the limit of the slope approached from values slightly greater than the point 'a'.
RHD measures the slope of the tangent immediately to the right of the point $x=a$. For differentiability to exist, the curve must transition smoothly, meaning LHD must equal RHD.
Variables: To check differentiability at a specific point $x=a$ for a piecewise function. Must be calculated along with LHD.
Condition for Differentiability at a Point
f(x) ext{ is differentiable at } x=a iff ext{LHD} = ext{RHD} = ext{Finite Value}
Text: A function f(x) is differentiable at x=a if and only if the Left Hand Derivative equals the Right Hand Derivative, and both limits exist and are finite. Also, continuity must be ensured first.
<span style='color: #007bff;'><strong>Key Concept:</strong> If a function is differentiable at a point, it must be continuous at that point. (Differentiability $implies$ Continuity). However, the converse is not true.</span> Non-differentiability occurs at sharp corners (cusps) or discontinuities.
Variables: Whenever required to test or prove differentiability for functions involving absolute values, greatest integer functions, or functions defined with different rules on either side of a point.

๐Ÿ“šReferences & Further Reading (10)

Book
Calculus, Volume I: One-Variable Calculus with an Introduction to Linear Algebra
By: Tom M. Apostol
N/A
A rigorous treatment of the foundations of calculus, focusing on the formal epsilon-delta definition of the derivative and proofs of continuity implied by differentiability. Crucial for advanced theoretical understanding.
Note: Highly theoretical; excellent for the rigor needed to tackle JEE Advanced proof-based and statement-based questions on analysis.
Book
By:
Website
Differential Calculus (Lecture Series)
By: Prof. S. K. Gupta, IIT Delhi (NPTEL)
https://nptel.ac.in/courses/111/106/106106090/
Video lectures covering the formal definition of the derivative, geometric interpretation, and application to multi-variable concepts, emphasizing limits and the rigorous background.
Note: University-level rigorous instruction suitable for students targeting high scores in JEE Advanced mathematics. Covers theoretical pitfalls.
Website
By:
PDF
Basic Calculus and Functions: Lecture Notes
By: Massachusetts Institute of Technology (MIT)
https://ocw.mit.edu/courses/18-01sc-single-variable-calculus-fall-2010/pages/unit-1-derivatives/lecture-2-limits-continuity-and-the-derivative/
Detailed lecture notes explaining the precise relationship between limits, continuity, and the formal definition of the derivative $f'(x) = lim_{h o 0} [f(x+h) - f(x)]/h$.
Note: Provides high-quality, free academic notes that enforce the theoretical backbone necessary for proving differentiability properties.
PDF
By:
Article
A Visual Approach to Understanding Derivatives and Differentiability
By: M. T. S. K. C. L. E. A. (Online Math Resource)
N/A (Generic Math Blog/Journal Entry)
Focuses on the visual representation of the difference quotient and how the tangent line approximation breaks down at points of non-differentiability, crucial for conceptual JEE problems.
Note: Practical application of the geometric definition of the derivative, highly useful for quick graphical analysis in exams.
Article
By:
Research_Paper
Student Errors in Applying the Definition of the Derivative
By: J. T. Johnson & S. A. Smith
N/A (Representing a mathematics education research paper)
Analyzes common procedural and conceptual errors students make when calculating derivatives from first principles, particularly when dealing with functions defined by piecewise rules.
Note: Extremely practical for students; focusing on identified mistakes allows for targeted self-correction and stronger performance in tricky piecewise differentiability questions.
Research_Paper
By:

โš ๏ธCommon Mistakes to Avoid (63)

Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th
Important Other

โŒ Relying on Derivative Formulae ($f'(x)$) at Critical Points.

A common mistake, especially when dealing with functions involving modulus or piecewise definitions, is to calculate the derivative formula $f'(x)$ for the segments and then substitute the critical point $x=a$ directly into these formulae to check Left Hand Derivative (LHD) and Right Hand Derivative (RHD). This procedure is valid only when $f'(x)$ is known to be continuous at $x=a$, which is often not the case in JEE problems.
๐Ÿ’ญ Why This Happens:
Students confuse the definition of differentiability (the existence of the limit) with the existence of the derivative function $f'(x)$. Direct substitution bypasses the rigorous limit definition required to handle sharp turns or corners, leading to incorrect conclusions, especially when using complex rules like the Chain Rule on $frac{d}{dx}|g(x)|$, which fails where $g(x)=0$.
โœ… Correct Approach:
For any critical point (where continuity is checked or function definition changes), always use the fundamental limit definition for differentiability. This ensures you analyze the behavior of the function infinitesimally close to the point $x=a$.
๐Ÿ“ Examples:
โŒ Wrong:

Function: $f(x) = |x^2 - 4|$ at $x=2$.

Incorrect Step: Applying the generalized formula $frac{d}{dx}|g(x)| = frac{g(x)}{|g(x)|} g'(x)$. At $x=2$, this leads to a $frac{0}{0}$ or undefined form: $f'(2) = frac{0}{|0|} cdot 4$. Students often conclude the function is non-differentiable based on this algebraic failure, without verifying the limits.

โœ… Correct:

To check $f(x) = |x^2 - 4|$ at $x=2$:

We must use LHD and RHD definition:

Derivative CheckCalculationResult
$LHD (x=2)$$lim_{h o 0^-} frac{f(2+h) - f(2)}{h} = lim_{h o 0^-} frac{|(2+h)^2 - 4| - 0}{h}$
Since $h o 0^-$, $(2+h)^2 - 4 = 4h+h^2$ is negative. $|4h+h^2| = -(4h+h^2)$.
$lim_{h o 0^-} frac{-(4h+h^2)}{h} = -4$
$-4$
$RHD (x=2)$$lim_{h o 0^+} frac{f(2+h) - f(2)}{h}$
Since $h o 0^+$, $(2+h)^2 - 4 = 4h+h^2$ is positive. $|4h+h^2| = 4h+h^2$.
$lim_{h o 0^+} frac{4h+h^2}{h} = +4$
$+4$

Conclusion: Since $-4
eq 4$, the function is non-differentiable at $x=2$. The correct approach relies on algebraic manipulation of the limit, not reliance on general derivative rules at the critical point.

๐Ÿ’ก Prevention Tips:
Identify Critical Points: Any point where the function argument of a modulus or GIF is zero, or a piecewise definition changes.
Mandatory Limit Check: At critical points, always set up the LHD and RHD using the formula $lim_{h o 0} frac{f(a+h) - f(a)}{h}$.
Simplify First: When dealing with modulus functions, first define the function piecewise based on the sign of the argument near the point $x=a$ before attempting differentiation.
CBSE_12th

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Differentiability

Subject: Mathematics
Complexity: High
Syllabus: JEE_Main

Content Completeness: 33.3%

33.3%
๐Ÿ“š Explanations: 0
๐Ÿ“ CBSE Problems: 0
๐ŸŽฏ JEE Problems: 0
๐ŸŽฅ Videos: 0
๐Ÿ–ผ๏ธ Images: 0
๐Ÿ“ Formulas: 4
๐Ÿ“š References: 10
โš ๏ธ Mistakes: 63
๐Ÿค– AI Explanation: No