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:
- The LHD at `x=a` exists and is finite.
- The RHD at `x=a` exists and is finite.
- 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 `ยฑโ`). 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 โค 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 โฅ 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 โค 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.