Hello there, aspiring mathematicians! Welcome to this crucial section where we'll unravel the mysteries of differentiating trigonometric and inverse trigonometric functions. Think of differentiation as finding the 'speed' at which a function's output changes relative to its input. We've seen it with polynomials, exponentials, and logarithms. Now, it's time to add our angular friends to the mix!

We're going to start from scratch, assuming you're just getting familiar with these. We'll build our understanding brick by brick, just like constructing a strong building. So, grab your notebooks and let's dive in!

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### Understanding the Basics: Why Do We Need These Rules?

Before we jump into the formulas, let's quickly recap what differentiation does for us. When we differentiate a function, say (y = f(x)), we find its derivative, denoted as (frac{dy}{dx}) or (f'(x)). This derivative tells us two incredibly important things:

1. Rate of Change: How quickly (y) changes as (x) changes. For example, if (x) is time and (y) is distance, (frac{dy}{dx}) is speed!
2. Slope of the Tangent: At any point on the curve of (y = f(x)), the derivative gives us the slope of the line tangent to the curve at that specific point. This is super useful for understanding the function's behavior.

Trigonometric functions like (sin(x)), (cos(x)), ( an(x)) are everywhere in physics (waves, oscillations), engineering (signal processing), and even computer graphics. Their rates of change are fundamental to understanding these phenomena. Similarly, inverse trigonometric functions like (sin^{-1}(x)) (also written as (arcsin(x))) are crucial when we need to find angles based on ratios.

So, let's learn how to find their 'speed' or 'slope'!

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### 1. Differentiating Trigonometric Functions

We have six primary trigonometric functions: sine, cosine, tangent, cotangent, secant, and cosecant. Each has a specific derivative that you'll need to remember and apply. Don't worry, there's a pattern, and with practice, they'll become second nature!

Let's list them out and then tackle some examples.

Function (f(x))	Derivative (f'(x)) or (frac{d}{dx}[f(x)])
(sin(x))	(mathbf{cos(x)})
(cos(x))	(mathbf{-sin(x)})
( an(x))	(mathbf{sec^2(x)})
(cot(x))	(mathbf{-csc^2(x)})
(sec(x))	(mathbf{sec(x) an(x)})
(csc(x))	(mathbf{-csc(x)cot(x)})

Quick Tip for Remembering the Signs: Notice that all the 'co-functions' ((cos(x)), (cot(x)), (csc(x))) have negative signs in their derivatives! This little trick can help you recall them faster.

Now, let's see these in action, especially with the Chain Rule, which is super important here. Remember, the chain rule states: if (y = f(u)) and (u = g(x)), then (frac{dy}{dx} = frac{dy}{du} cdot frac{du}{dx}). In simpler terms, differentiate the 'outer' function first, then multiply by the derivative of the 'inner' function.

#### Example 1: Basic Application

Find the derivative of (y = sin(x) + cos(x) - an(x)).

Solution:
Using the sum/difference rule for differentiation, we differentiate each term separately:
(frac{dy}{dx} = frac{d}{dx}[sin(x)] + frac{d}{dx}[cos(x)] - frac{d}{dx}[ an(x)])
(frac{dy}{dx} = cos(x) + (-sin(x)) - sec^2(x))
(mathbf{frac{dy}{dx} = cos(x) - sin(x) - sec^2(x)})

#### Example 2: Applying the Chain Rule

Find the derivative of (f(x) = sin(2x)).

Solution:
Here, the outer function is (sin(u)) and the inner function is (u = 2x).
1. Differentiate the outer function ((sin(u))) with respect to (u): (frac{d}{du}[sin(u)] = cos(u)).
2. Substitute (u = 2x) back: (cos(2x)).
3. Differentiate the inner function ((2x)) with respect to (x): (frac{d}{dx}[2x] = 2).
4. Multiply the results: (mathbf{f'(x) = cos(2x) cdot 2 = 2cos(2x)}).

#### Example 3: More Chain Rule Fun!

Find the derivative of (y = an(x^2 + 3)).

Solution:
Outer function: ( an(u)) where (u = x^2 + 3).
1. Derivative of outer: (frac{d}{du}[ an(u)] = sec^2(u)).
2. Substitute (u = x^2 + 3): (sec^2(x^2 + 3)).
3. Derivative of inner: (frac{d}{dx}[x^2 + 3] = 2x + 0 = 2x).
4. Multiply: (mathbf{frac{dy}{dx} = sec^2(x^2 + 3) cdot (2x) = 2xsec^2(x^2 + 3)}).

#### Example 4: Product Rule with Trig Functions

Find the derivative of (f(x) = x^3 cos(x)).

Solution:
This is a product of two functions, (u = x^3) and (v = cos(x)).
Recall the product rule: ((uv)' = u'v + uv').
1. (u = x^3 implies u' = 3x^2)
2. (v = cos(x) implies v' = -sin(x))
3. Apply the product rule:
(f'(x) = (3x^2)(cos(x)) + (x^3)(-sin(x)))
(mathbf{f'(x) = 3x^2cos(x) - x^3sin(x)})

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### 2. Differentiating Inverse Trigonometric Functions

Inverse trigonometric functions answer the question: "What angle has this sine/cosine/tangent value?" For example, (sin^{-1}(0.5)) asks for the angle whose sine is 0.5 (which is (pi/6) or 30 degrees).

Their derivatives are a bit more complex than the regular trig functions, and they almost always involve square roots. These formulas are absolutely essential for JEE, so commit them to memory!

Function (f(x))	Derivative (f'(x)) or (frac{d}{dx}[f(x)])	Domain for (x)
(sin^{-1}(x))	(mathbf{frac{1}{sqrt{1-x^2}}})	(-1 < x < 1)
(cos^{-1}(x))	(mathbf{frac{-1}{sqrt{1-x^2}}})	(-1 < x < 1)
( an^{-1}(x))	(mathbf{frac{1}{1+x^2}})	All real (x)
(cot^{-1}(x))	(mathbf{frac{-1}{1+x^2}})	All real (x)
(sec^{-1}(x))	(mathbf{frac{1}{|x|sqrt{x^2-1}}})	(|x| > 1)
(csc^{-1}(x))	(mathbf{frac{-1}{|x|sqrt{x^2-1}}})	(|x| > 1)

Quick Tip for Remembering the Signs: Just like with regular trig functions, the derivatives of the 'co-inverse functions' ((cos^{-1}(x)), (cot^{-1}(x)), (csc^{-1}(x))) are the negative of their non-co counterparts. This makes memorizing much easier!

Let's apply these with some examples. Again, the Chain Rule is your best friend here!

#### Example 5: Basic Inverse Trig Derivative

Find the derivative of (y = sin^{-1}(x) + an^{-1}(x)).

Solution:
(frac{dy}{dx} = frac{d}{dx}[sin^{-1}(x)] + frac{d}{dx}[ an^{-1}(x)])
(mathbf{frac{dy}{dx} = frac{1}{sqrt{1-x^2}} + frac{1}{1+x^2}})

#### Example 6: Chain Rule with Inverse Sine

Find the derivative of (f(x) = sin^{-1}(2x)).

Solution:
Outer function: (sin^{-1}(u)) where (u = 2x).
1. Derivative of outer: (frac{d}{du}[sin^{-1}(u)] = frac{1}{sqrt{1-u^2}}).
2. Substitute (u = 2x): (frac{1}{sqrt{1-(2x)^2}} = frac{1}{sqrt{1-4x^2}}).
3. Derivative of inner: (frac{d}{dx}[2x] = 2).
4. Multiply: (mathbf{f'(x) = frac{1}{sqrt{1-4x^2}} cdot 2 = frac{2}{sqrt{1-4x^2}}}).

#### Example 7: Chain Rule with Inverse Tangent and an Exponential

Find the derivative of (y = an^{-1}(e^x)).

Solution:
Outer function: ( an^{-1}(u)) where (u = e^x).
1. Derivative of outer: (frac{d}{du}[ an^{-1}(u)] = frac{1}{1+u^2}).
2. Substitute (u = e^x): (frac{1}{1+(e^x)^2} = frac{1}{1+e^{2x}}).
3. Derivative of inner: (frac{d}{dx}[e^x] = e^x).
4. Multiply: (mathbf{frac{dy}{dx} = frac{1}{1+e^{2x}} cdot e^x = frac{e^x}{1+e^{2x}}}).

#### Example 8: Product Rule with Inverse Cosine

Find the derivative of (g(x) = x^2 cos^{-1}(x)).

Solution:
This is a product of (u = x^2) and (v = cos^{-1}(x)).
1. (u = x^2 implies u' = 2x)
2. (v = cos^{-1}(x) implies v' = frac{-1}{sqrt{1-x^2}})
3. Apply the product rule:
(g'(x) = (2x)(cos^{-1}(x)) + (x^2)left(frac{-1}{sqrt{1-x^2}}
ight))
(mathbf{g'(x) = 2xcos^{-1}(x) - frac{x^2}{sqrt{1-x^2}}})

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### 3. CBSE vs. JEE Focus

For CBSE exams, knowing these formulas and applying the chain rule, product rule, and quotient rule correctly is usually sufficient. Questions tend to be direct applications.

For JEE Mains & Advanced, these formulas are just the starting point. You'll need to master their application in more complex scenarios:
* Nested Chain Rule: Differentiating functions like (sin(cos(e^{x^2}))).
* Implicit Differentiation: When (x) and (y) are mixed in equations like (sin(xy) = x^2y).
* Logarithmic Differentiation: For functions like ((sin x)^x).
* Inverse Functions: Finding derivatives of functions like (f^{-1}(x)) using the formula ((f^{-1})'(x) = frac{1}{f'(f^{-1}(x))}).
* Substitutions & Simplification: Often, especially with inverse trig functions, you'll need to simplify the expression using trigonometric identities before differentiating. For example, differentiating ( an^{-1}left(frac{2x}{1-x^2}
ight)) is much easier if you first substitute (x = an heta). We'll explore these deeper in the 'deep_dive' section.

The key takeaway is that for JEE, you need to be comfortable combining these rules with everything else you've learned about differentiation and algebraic manipulation.

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### Recap and What's Next?

You've now got a solid foundation for differentiating trigonometric and inverse trigonometric functions! We've covered:

* The derivatives of all six basic trigonometric functions.
* The derivatives of all six basic inverse trigonometric functions.
* How to apply the Chain Rule, Product Rule, and Quotient Rule in conjunction with these new derivatives.
* The importance of memorizing these formulas for both CBSE and JEE.

Practice is your best friend here. The more you work through examples, the more natural these formulas and their applications will feel. In the upcoming sections, we'll delve even deeper, exploring the derivations of these formulas, tackling more advanced problems, and looking at specific JEE-level strategies. Keep up the great work!

Welcome, future engineers! In this deep dive, we're going to rigorously explore one of the most fundamental aspects of differential calculus: the differentiation of trigonometric and inverse trigonometric functions. This is not just about memorizing formulas; it's about understanding their derivations, applying them systematically, and mastering advanced techniques crucial for cracking the JEE. So, grab your notebooks, and let's embark on this journey!

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1. The Foundation: Derivatives of Basic Trigonometric Functions

Before we jump into the specific functions, let's quickly recap our basic differentiation toolkit: the sum rule, product rule, quotient rule, and most importantly, the chain rule. These will be our constant companions.

1.1 Derivations from First Principles (or Limit Definition)

Recall that the derivative of a function $f(x)$ is defined as:
$f'(x) = lim_{h o 0} frac{f(x+h) - f(x)}{h}$

Let's use this to derive the derivative of $sin x$.

1. Derivative of $sin x$:

Let $f(x) = sin x$.
$f'(x) = lim_{h o 0} frac{sin(x+h) - sin x}{h}$
Using the sum-to-product formula: $sin A - sin B = 2 cos left(frac{A+B}{2}
ight) sin left(frac{A-B}{2}
ight)$
Here, $A = x+h$ and $B = x$.
$A+B = 2x+h$, so $frac{A+B}{2} = x + frac{h}{2}$
$A-B = h$, so $frac{A-B}{2} = frac{h}{2}$
$f'(x) = lim_{h o 0} frac{2 cosleft(x + frac{h}{2}
ight) sinleft(frac{h}{2}
ight)}{h}$
We can rewrite this as:
$f'(x) = lim_{h o 0} left[ cosleft(x + frac{h}{2}
ight) cdot frac{sinleft(frac{h}{2}
ight)}{frac{h}{2}}
ight]$
As $h o 0$, we know that $lim_{ heta o 0} frac{sin heta}{ heta} = 1$. Here $ heta = frac{h}{2}$.
Also, $lim_{h o 0} cosleft(x + frac{h}{2}
ight) = cos x$.
Therefore, $mathbf{frac{d}{dx}(sin x) = cos x}$.


2. Derivative of $cos x$:

Similarly, for $f(x) = cos x$:
$f'(x) = lim_{h o 0} frac{cos(x+h) - cos x}{h}$
Using the sum-to-product formula: $cos A - cos B = -2 sin left(frac{A+B}{2}
ight) sin left(frac{A-B}{2}
ight)$
$f'(x) = lim_{h o 0} frac{-2 sinleft(x + frac{h}{2}
ight) sinleft(frac{h}{2}
ight)}{h}$
$f'(x) = lim_{h o 0} left[ -sinleft(x + frac{h}{2}
ight) cdot frac{sinleft(frac{h}{2}
ight)}{frac{h}{2}}
ight]$
As $h o 0$, this becomes $-sin x cdot 1$.
Therefore, $mathbf{frac{d}{dx}(cos x) = -sin x}$.

1.2 Derivatives of Other Trigonometric Functions (using Quotient Rule)

The remaining trigonometric derivatives can be found by expressing them in terms of $sin x$ and $cos x$ and applying the quotient rule:
If $y = frac{u(x)}{v(x)}$, then $frac{dy}{dx} = frac{u'v - uv'}{v^2}$.

1. Derivative of $ an x$:

Since $ an x = frac{sin x}{cos x}$, let $u = sin x$ and $v = cos x$.
Then $u' = cos x$ and $v' = -sin x$.
$frac{d}{dx}( an x) = frac{(cos x)(cos x) - (sin x)(-sin x)}{(cos x)^2}$
$= frac{cos^2 x + sin^2 x}{cos^2 x} = frac{1}{cos^2 x} = mathbf{sec^2 x}$.


2. Derivative of $cot x$:

Since $cot x = frac{cos x}{sin x}$, let $u = cos x$ and $v = sin x$.
Then $u' = -sin x$ and $v' = cos x$.
$frac{d}{dx}(cot x) = frac{(-sin x)(sin x) - (cos x)(cos x)}{(sin x)^2}$
$= frac{-sin^2 x - cos^2 x}{sin^2 x} = frac{-(sin^2 x + cos^2 x)}{sin^2 x} = frac{-1}{sin^2 x} = mathbf{-csc^2 x}$.


3. Derivative of $sec x$:

Since $sec x = frac{1}{cos x}$, let $u = 1$ and $v = cos x$.
Then $u' = 0$ and $v' = -sin x$.
$frac{d}{dx}(sec x) = frac{(0)(cos x) - (1)(-sin x)}{(cos x)^2}$
$= frac{sin x}{cos^2 x} = frac{1}{cos x} cdot frac{sin x}{cos x} = mathbf{sec x an x}$.


4. Derivative of $csc x$:

Since $csc x = frac{1}{sin x}$, let $u = 1$ and $v = sin x$.
Then $u' = 0$ and $v' = cos x$.
$frac{d}{dx}(csc x) = frac{(0)(sin x) - (1)(cos x)}{(sin x)^2}$
$= frac{-cos x}{sin^2 x} = -frac{1}{sin x} cdot frac{cos x}{sin x} = mathbf{-csc x cot x}$.

1.3 Summary of Basic Trigonometric Derivatives

It's crucial to memorize these, but understand their origin.

Function $f(x)$	Derivative $f'(x)$
$sin x$	$cos x$
$cos x$	$-sin x$
$ an x$	$sec^2 x$
$cot x$	$-csc^2 x$
$sec x$	$sec x an x$
$csc x$	$-csc x cot x$

1.4 Examples with Chain Rule

The Chain Rule is indispensable here. If $y = f(g(x))$, then $frac{dy}{dx} = f'(g(x)) cdot g'(x)$.

Example 1.1: Differentiate $y = sin(x^2 + 5)$.

Solution:

Let $u = x^2 + 5$. Then $y = sin u$.
Using the chain rule: $frac{dy}{dx} = frac{dy}{du} cdot frac{du}{dx}$.
$frac{dy}{du} = frac{d}{du}(sin u) = cos u$.
$frac{du}{dx} = frac{d}{dx}(x^2 + 5) = 2x$.
So, $frac{dy}{dx} = (cos u) cdot (2x) = 2x cos(x^2 + 5)$.

Example 1.2: Find $frac{dy}{dx}$ for $y = sec^3(4x)$.

Solution:

This can be written as $y = (sec(4x))^3$. This involves a power function, then a secant function, then a linear function.
Let $u = sec(4x)$. Then $y = u^3$.
$frac{dy}{du} = 3u^2 = 3(sec(4x))^2 = 3sec^2(4x)$.
Now, we need $frac{du}{dx} = frac{d}{dx}(sec(4x))$.
Let $v = 4x$. Then $u = sec v$.
$frac{du}{dv} = sec v an v = sec(4x) an(4x)$.
$frac{dv}{dx} = frac{d}{dx}(4x) = 4$.
So, $frac{du}{dx} = frac{du}{dv} cdot frac{dv}{dx} = (sec(4x) an(4x)) cdot 4 = 4sec(4x) an(4x)$.
Finally, combining everything:
$frac{dy}{dx} = frac{dy}{du} cdot frac{du}{dx} = (3sec^2(4x)) cdot (4sec(4x) an(4x))$
$frac{dy}{dx} = 12sec^3(4x) an(4x)$.

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2. The Inverse World: Derivatives of Inverse Trigonometric Functions

The derivatives of inverse trigonometric functions are derived using a powerful technique called implicit differentiation. The key idea is to "undo" the inverse function by applying the original trigonometric function to both sides.

2.1 Derivations Using Implicit Differentiation

1. Derivative of $sin^{-1} x$:

Let $y = sin^{-1} x$.
This implies $sin y = x$. (Remember: $y$ is an angle whose sine is $x$.)
Now, differentiate both sides with respect to $x$:
$frac{d}{dx}(sin y) = frac{d}{dx}(x)$
Using the chain rule on the left side: $cos y cdot frac{dy}{dx} = 1$.
So, $frac{dy}{dx} = frac{1}{cos y}$.
We need to express $cos y$ in terms of $x$. Since $sin y = x$, and $sin^2 y + cos^2 y = 1$, we have $cos^2 y = 1 - sin^2 y = 1 - x^2$.
Thus, $cos y = pmsqrt{1 - x^2}$.
For $y = sin^{-1} x$, the range is $[-frac{pi}{2}, frac{pi}{2}]$. In this range, $cos y ge 0$.
So, $cos y = sqrt{1 - x^2}$.
Therefore, $mathbf{frac{d}{dx}(sin^{-1} x) = frac{1}{sqrt{1 - x^2}}}$ (for $-1 < x < 1$).


Note: The derivative is undefined at $x=pm 1$ because the tangent to the curve $y=sin^{-1} x$ is vertical at these points.




2. Derivative of $cos^{-1} x$:

Let $y = cos^{-1} x$.
This implies $cos y = x$.
Differentiate both sides with respect to $x$:
$frac{d}{dx}(cos y) = frac{d}{dx}(x)$
$-sin y cdot frac{dy}{dx} = 1$.
So, $frac{dy}{dx} = -frac{1}{sin y}$.
Since $cos y = x$, and $sin^2 y + cos^2 y = 1$, we have $sin^2 y = 1 - cos^2 y = 1 - x^2$.
Thus, $sin y = pmsqrt{1 - x^2}$.
For $y = cos^{-1} x$, the range is $[0, pi]$. In this range, $sin y ge 0$.
So, $sin y = sqrt{1 - x^2}$.
Therefore, $mathbf{frac{d}{dx}(cos^{-1} x) = -frac{1}{sqrt{1 - x^2}}}$ (for $-1 < x < 1$).


3. Derivative of $ an^{-1} x$:

Let $y = an^{-1} x$.
This implies $ an y = x$.
Differentiate both sides with respect to $x$:
$frac{d}{dx}( an y) = frac{d}{dx}(x)$
$sec^2 y cdot frac{dy}{dx} = 1$.
So, $frac{dy}{dx} = frac{1}{sec^2 y}$.
Using the identity $sec^2 y = 1 + an^2 y$:
$sec^2 y = 1 + x^2$.
Therefore, $mathbf{frac{d}{dx}( an^{-1} x) = frac{1}{1 + x^2}}$.


4. Derivative of $cot^{-1} x$:

Let $y = cot^{-1} x$.
This implies $cot y = x$.
Differentiate both sides with respect to $x$:
$frac{d}{dx}(cot y) = frac{d}{dx}(x)$
$-csc^2 y cdot frac{dy}{dx} = 1$.
So, $frac{dy}{dx} = -frac{1}{csc^2 y}$.
Using the identity $csc^2 y = 1 + cot^2 y$:
$csc^2 y = 1 + x^2$.
Therefore, $mathbf{frac{d}{dx}(cot^{-1} x) = -frac{1}{1 + x^2}}$.


5. Derivative of $sec^{-1} x$:

Let $y = sec^{-1} x$.
This implies $sec y = x$.
Differentiate both sides with respect to $x$:
$frac{d}{dx}(sec y) = frac{d}{dx}(x)$
$sec y an y cdot frac{dy}{dx} = 1$.
So, $frac{dy}{dx} = frac{1}{sec y an y}$.
We know $sec y = x$. We need $ an y$ in terms of $x$.
Using the identity $ an^2 y = sec^2 y - 1$:
$ an^2 y = x^2 - 1$, so $ an y = pmsqrt{x^2 - 1}$.
For $y = sec^{-1} x$, the range is $[0, frac{pi}{2}) cup (frac{pi}{2}, pi]$.



• If $x > 1$ (i.e., $y in [0, frac{pi}{2})$), then $ an y > 0$, so $ an y = sqrt{x^2 - 1}$.


• If $x < -1$ (i.e., $y in (frac{pi}{2}, pi]$), then $ an y < 0$, so $ an y = -sqrt{x^2 - 1}$.



We can combine these by writing $ an y = ext{sgn}(x)sqrt{x^2-1}$ or more commonly, noting that $sec y = x$, we have $x sqrt{x^2-1}$ if $ an y$ is positive, and $x (-sqrt{x^2-1})$ if $ an y$ is negative.
A more elegant way to handle the sign is to use $|x|$:
Since $x = sec y$, $|x| = |sec y|$.
$frac{dy}{dx} = frac{1}{x an y} = frac{1}{x (pmsqrt{x^2-1})}$.
The sign of $ an y$ is the same as the sign of $x$ when considering the principal value range for $sec^{-1}x$.
Thus, $frac{dy}{dx} = mathbf{frac{1}{|x|sqrt{x^2 - 1}}}$ (for $|x| > 1$).


Note: The absolute value $|x|$ is crucial here. If we used $xsqrt{x^2-1}$ directly, for $x < -1$, $x$ would be negative, but $ an y$ (for $y in (pi/2, pi]$) is also negative, making the product positive, consistent with the formula.




6. Derivative of $csc^{-1} x$:

Let $y = csc^{-1} x$.
This implies $csc y = x$.
Differentiate both sides with respect to $x$:
$frac{d}{dx}(csc y) = frac{d}{dx}(x)$
$-csc y cot y cdot frac{dy}{dx} = 1$.
So, $frac{dy}{dx} = -frac{1}{csc y cot y}$.
We know $csc y = x$. We need $cot y$ in terms of $x$.
Using the identity $cot^2 y = csc^2 y - 1$:
$cot^2 y = x^2 - 1$, so $cot y = pmsqrt{x^2 - 1}$.
For $y = csc^{-1} x$, the range is $[-frac{pi}{2}, 0) cup (0, frac{pi}{2}]$.
The sign of $cot y$ is the same as the sign of $x$ in this principal value range.
Therefore, $mathbf{frac{d}{dx}(csc^{-1} x) = -frac{1}{|x|sqrt{x^2 - 1}}}$ (for $|x| > 1$).

2.2 Summary of Inverse Trigonometric Derivatives

Function $f(x)$	Derivative $f'(x)$	Domain of Derivative
$sin^{-1} x$	$frac{1}{sqrt{1 - x^2}}$	$(-1, 1)$
$cos^{-1} x$	$-frac{1}{sqrt{1 - x^2}}$	$(-1, 1)$
$ an^{-1} x$	$frac{1}{1 + x^2}$	$(-infty, infty)$
$cot^{-1} x$	$-frac{1}{1 + x^2}$	$(-infty, infty)$
$sec^{-1} x$	$frac{1}{|x|sqrt{x^2 - 1}}$	$(-infty, -1) cup (1, infty)$
$csc^{-1} x$	$-frac{1}{|x|sqrt{x^2 - 1}}$	$(-infty, -1) cup (1, infty)$

2.3 Examples with Chain Rule

Example 2.1: Differentiate $y = an^{-1}(ln x)$.

Solution:

Let $u = ln x$. Then $y = an^{-1} u$.
Using the chain rule: $frac{dy}{dx} = frac{dy}{du} cdot frac{du}{dx}$.
$frac{dy}{du} = frac{d}{du}( an^{-1} u) = frac{1}{1 + u^2}$.
$frac{du}{dx} = frac{d}{dx}(ln x) = frac{1}{x}$.
So, $frac{dy}{dx} = frac{1}{1 + (ln x)^2} cdot frac{1}{x} = frac{1}{x(1 + (ln x)^2)}$.

Example 2.2: Find $frac{dy}{dx}$ for $y = sin^{-1}(2xsqrt{1-x^2})$.

Solution:

This function looks complicated. A direct application of the chain rule would be tedious.
Let $u = 2xsqrt{1-x^2}$. Then $y = sin^{-1} u$.
$frac{dy}{du} = frac{1}{sqrt{1-u^2}} = frac{1}{sqrt{1-(2xsqrt{1-x^2})^2}} = frac{1}{sqrt{1-4x^2(1-x^2)}} = frac{1}{sqrt{1-4x^2+4x^4}}$.
This looks like $sqrt{(1-2x^2)^2}$ but be careful with signs. $1-4x^2+4x^4 = (1-2x^2)^2$.
So $frac{dy}{du} = frac{1}{|1-2x^2|}$.
Now $frac{du}{dx} = frac{d}{dx}(2xsqrt{1-x^2}) = 2frac{d}{dx}(xsqrt{1-x^2})$.
Using product rule on $xsqrt{1-x^2}$:
$frac{d}{dx}(xsqrt{1-x^2}) = 1 cdot sqrt{1-x^2} + x cdot frac{1}{2sqrt{1-x^2}} cdot (-2x)$
$= sqrt{1-x^2} - frac{x^2}{sqrt{1-x^2}} = frac{1-x^2-x^2}{sqrt{1-x^2}} = frac{1-2x^2}{sqrt{1-x^2}}$.
So, $frac{du}{dx} = 2 frac{1-2x^2}{sqrt{1-x^2}}$.
Finally, $frac{dy}{dx} = frac{1}{|1-2x^2|} cdot 2 frac{1-2x^2}{sqrt{1-x^2}}$.
If $1-2x^2 > 0$, $frac{dy}{dx} = frac{1}{1-2x^2} cdot 2 frac{1-2x^2}{sqrt{1-x^2}} = frac{2}{sqrt{1-x^2}}$.
If $1-2x^2 < 0$, $frac{dy}{dx} = frac{1}{-(1-2x^2)} cdot 2 frac{1-2x^2}{sqrt{1-x^2}} = -frac{2}{sqrt{1-x^2}}$.
This piecewise result indicates that there's a simpler way!

---

3. JEE Advanced Strategies: Simplification before Differentiation

The previous example illustrates a critical JEE strategy: simplify the expression using properties of inverse trigonometric functions BEFORE differentiating. This often involves a trigonometric substitution.

3.1 Trigonometric Substitutions

Often, expressions like $sqrt{a^2-x^2}$, $sqrt{a^2+x^2}$, $sqrt{x^2-a^2}$ or rational functions appear inside inverse trigonometric functions. These are strong indicators for substitution.

Expression	Suggested Substitution	Resulting Identity
$sqrt{a^2 - x^2}$	$x = asin heta$ or $x = acos heta$	$sqrt{a^2 - a^2sin^2 heta} = a|cos heta|$
$sqrt{a^2 + x^2}$	$x = a an heta$ or $x = acot heta$	$sqrt{a^2 + a^2 an^2 heta} = a|sec heta|$
$sqrt{x^2 - a^2}$	$x = asec heta$ or $x = acsc heta$	$sqrt{a^2sec^2 heta - a^2} = a| an heta|$
$frac{2x}{1+x^2}$, $frac{1-x^2}{1+x^2}$	$x = an heta$	Double angle formulas for $sin(2 heta), cos(2 heta)$
$3x-4x^3$, $4x^3-3x$	$x = sin heta$	Triple angle formulas for $sin(3 heta)$
$frac{3x-x^3}{1-3x^2}$	$x = an heta$	Triple angle formulas for $ an(3 heta)$

Let's re-evaluate Example 2.2 using substitution.

Example 3.1: Find $frac{dy}{dx}$ for $y = sin^{-1}(2xsqrt{1-x^2})$.

Solution:

Notice the term $sqrt{1-x^2}$. This suggests $x = sin heta$.
Let $x = sin heta$. Then $ heta = sin^{-1} x$.
Substitute into $y$:
$y = sin^{-1}(2sin hetasqrt{1-sin^2 heta})$
$y = sin^{-1}(2sin hetasqrt{cos^2 heta})$
$y = sin^{-1}(2sin heta|cos heta|)$.
Crucial Step: Domain Consideration. The domain of $sin^{-1}(2xsqrt{1-x^2})$ requires $-1 le 2xsqrt{1-x^2} le 1$. Also, for $sqrt{1-x^2}$ to be real, $-1 le x le 1$.
If we assume the principal value branch for $sin^{-1} x$, i.e., $-frac{pi}{2} le heta le frac{pi}{2}$, then $cos heta ge 0$.
So, $|cos heta| = cos heta$.
$y = sin^{-1}(2sin hetacos heta)$
$y = sin^{-1}(sin(2 heta))$.

Now, remember the property: $sin^{-1}(sin A) = A$ only if $A$ lies in the principal value range of $sin^{-1}$, i.e., $[-frac{pi}{2}, frac{pi}{2}]$.
Since $-frac{pi}{2} le heta le frac{pi}{2}$, we have $-pi le 2 heta le pi$.
So we need to consider cases:

• Case 1: If $-frac{pi}{4} le heta le frac{pi}{4}$ (i.e., $-frac{1}{sqrt{2}} le x le frac{1}{sqrt{2}}$), then $-frac{pi}{2} le 2 heta le frac{pi}{2}$.
  In this case, $y = 2 heta = 2sin^{-1} x$.
  $frac{dy}{dx} = 2 cdot frac{1}{sqrt{1-x^2}}$.


• Case 2: If $frac{pi}{4} < heta le frac{pi}{2}$ (i.e., $frac{1}{sqrt{2}} < x le 1$), then $frac{pi}{2} < 2 heta le pi$.
  In this case, $sin^{-1}(sin(2 heta)) = pi - 2 heta$.
  So, $y = pi - 2 heta = pi - 2sin^{-1} x$.
  $frac{dy}{dx} = 0 - 2 cdot frac{1}{sqrt{1-x^2}} = -frac{2}{sqrt{1-x^2}}$.


• Case 3: If $-frac{pi}{2} le heta < -frac{pi}{4}$ (i.e., $-1 le x < -frac{1}{sqrt{2}}$), then $-pi le 2 heta < -frac{pi}{2}$.
  In this case, $sin^{-1}(sin(2 heta)) = -pi - 2 heta$ (using property $sin^{-1}(sin x) = -pi - x$ if $x in [-pi, -pi/2]$).
  So, $y = -pi - 2 heta = -pi - 2sin^{-1} x$.
  $frac{dy}{dx} = 0 - 2 cdot frac{1}{sqrt{1-x^2}} = -frac{2}{sqrt{1-x^2}}$.

This shows how important it is to be careful with the range of the principal values. This level of detail is a JEE Advanced focus. For CBSE, often the domain is implicitly chosen to be the simpler case.

Example 3.2: Differentiate $y = an^{-1}left(frac{3x - x^3}{1 - 3x^2}
ight)$, assuming $-frac{1}{sqrt{3}} < x < frac{1}{sqrt{3}}$.

Solution:

The expression inside $ an^{-1}$ looks very familiar! It's the formula for $ an(3 heta)$.
Let $x = an heta$. Then $ heta = an^{-1} x$.
The condition $-frac{1}{sqrt{3}} < x < frac{1}{sqrt{3}}$ implies $-frac{pi}{6} < heta < frac{pi}{6}$.
Substitute $x = an heta$ into $y$:
$y = an^{-1}left(frac{3 an heta - an^3 heta}{1 - 3 an^2 heta}
ight)$
$y = an^{-1}( an(3 heta))$.
Since $-frac{pi}{6} < heta < frac{pi}{6}$, we have $-frac{pi}{2} < 3 heta < frac{pi}{2}$.
This means $3 heta$ lies in the principal value range of $ an^{-1}$.
So, $y = 3 heta = 3 an^{-1} x$.
Now differentiate:
$frac{dy}{dx} = 3 cdot frac{d}{dx}( an^{-1} x) = mathbf{frac{3}{1 + x^2}}$.
This simplification is a massive time-saver and accuracy booster compared to direct differentiation using the chain rule multiple times.

3.2 Logarithmic Differentiation with Trigonometric Functions

When a function is in the form of $f(x)^{g(x)}$ (e.g., $(sin x)^x$), or involves complex products/quotients, logarithmic differentiation is often the easiest approach.

Example 3.3: Find $frac{dy}{dx}$ for $y = (sin x)^{cos x}$.

Solution:

Take natural logarithm on both sides:
$ln y = ln((sin x)^{cos x})$
$ln y = cos x cdot ln(sin x)$.
Now, differentiate both sides with respect to $x$:
$frac{1}{y} frac{dy}{dx} = frac{d}{dx}(cos x cdot ln(sin x))$.
Using the product rule on the right side ($u = cos x, v = ln(sin x)$):
$u' = -sin x$.
$v' = frac{d}{dx}(ln(sin x)) = frac{1}{sin x} cdot cos x = cot x$.
So, $frac{1}{y} frac{dy}{dx} = (-sin x)(ln(sin x)) + (cos x)(cot x)$
$frac{1}{y} frac{dy}{dx} = -sin x ln(sin x) + cos x frac{cos x}{sin x}$
$frac{1}{y} frac{dy}{dx} = -sin x ln(sin x) + frac{cos^2 x}{sin x}$.
Finally, multiply by $y$:
$frac{dy}{dx} = y left( -sin x ln(sin x) + frac{cos^2 x}{sin x}
ight)$
$frac{dy}{dx} = (sin x)^{cos x} left( -sin x ln(sin x) + cot x cos x
ight)$.

---

4. CBSE vs. JEE Focus

• CBSE Focus: The emphasis is primarily on memorizing the derivative formulas for trigonometric and inverse trigonometric functions and applying the chain rule, product rule, and quotient rule directly. Simplification using basic trigonometric identities is expected, but the nuanced handling of inverse trigonometric function properties (like the range of $sin^{-1}(sin A)$) is often simplified or tested in specific domains that avoid complications.


• JEE Focus: While the basic formulas are essential, JEE goes significantly deeper.
  
  
  
  • Simplification is Key: Recognizing expressions that can be simplified using trigonometric substitutions (e.g., $x= an heta, x=sin heta$) before differentiation is critical. This transforms complex problems into much simpler ones.
  
  
  • Domain and Range: Thorough understanding of the domain and range of inverse trigonometric functions, especially their principal values, is paramount. Problems like Example 3.1, where $sin^{-1}(sin(2 heta))$ needs careful handling based on the value of $2 heta$, are typical in JEE. Ignoring this can lead to incorrect or incomplete answers.
  
  
  • Implicit and Logarithmic Differentiation: These techniques are heavily tested in conjunction with trigonometric and inverse trigonometric functions, sometimes in combination with other function types.
  
  
  • Higher Order Derivatives: Expect questions involving finding second or third derivatives of these functions, often after an initial simplification.
  
  
  
  

---

Key Takeaways

1. Master the six basic trigonometric derivatives and their inverse counterparts. Understand their derivations using first principles and implicit differentiation.


2. The Chain Rule is your most powerful tool for differentiating composite functions involving these.


3. For inverse trigonometric functions, always look for opportunities to simplify the expression first using trigonometric substitutions and inverse trigonometric properties. This is a common JEE strategy.


4. Pay close attention to the domain and range restrictions when simplifying expressions like $sin^{-1}(sin heta)$ or $ an^{-1}( an heta)$. This is where many students make mistakes in JEE.


5. Be proficient in applying logarithmic differentiation when dealing with functions like $f(x)^{g(x)}$ or complex products/quotients.

By thoroughly understanding these concepts and practicing with a variety of problems, you'll be well-prepared to tackle any differentiation problem involving trigonometric and inverse trigonometric functions that JEE throws your way! Keep practicing, and you'll find these challenging concepts becoming intuitive.

Mastering the derivatives of trigonometric and inverse trigonometric functions is fundamental for JEE and board exams. While direct memorization works, mnemonics and practical shortcuts can significantly boost recall speed and accuracy, especially under exam pressure.

Mnemonics for Trigonometric Derivatives

The six basic trigonometric derivatives can be easily recalled using these simple tricks:

• 
  The "Co" Rule for Negative Signs:
  

  Mnemonic: "If it starts with 'Co', the derivative goes low (negative)."

  
  

  This means derivatives of cos x, cot x, and cosec x will always have a negative sign.

  
  
  
  
  • d/dx (sin x) = cos x
  
  
  • d/dx (cos x) = -sin x
  
  
  • d/dx (tan x) = sec² x
  
  
  • d/dx (cot x) = -cosec² x
  
  
  • d/dx (sec x) = sec x tan x
  
  
  • d/dx (cosec x) = -cosec x cot x
  
  
  
  


• 
  Derivative Pairing & Transformations:
  
  
  
  • 
    "Swap Twins": sin x and cos x are like twins.
    
    
    
    • (sin x)' = cos x
    
    
    • (cos x)' = -sin x (The 'co' rule applies!)
    
    
    
    
  
  
  • 
    "Square Cousins": tan x and cot x give squared terms.
    
    
    
    • (tan x)' = sec² x (Think 'tan' is related to 'sec')
    
    
    • (cot x)' = -cosec² x (Again, 'co' rule for negative)
    
    
    
    
  
  
  • 
    "Self-Partner Duos": sec x and cosec x derivatives involve themselves and another trig function.
    
    
    
    • (sec x)' = sec x tan x (sec and tan are partners)
    
    
    • (cosec x)' = -cosec x cot x ('co' rule, cosec and cot are partners)
    
    
    
    
  
  
  
  

Mnemonics for Inverse Trigonometric Derivatives

The inverse trigonometric derivatives follow similar patterns:

• 
  The "Co" Rule Extends:
  

  Mnemonic: "Inverse 'Co' functions also carry a 'No' (negative) sign."

  
  

  Derivatives of cos⁻¹ x, cot⁻¹ x, and cosec⁻¹ x are negative.

  
  


• 
  Denominator Family Mnemonics:
  

  Focus on the denominator structure, which forms natural groupings:

  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  

  Inverse Trig Function	Derivative	Mnemonic/Denominator Family
  sin⁻¹ x	1 / √(1 - x²)	"Root of One Minus X-Squared" Family
  cos⁻¹ x	-1 / √(1 - x²)
  tan⁻¹ x	1 / (1 + x²)	"One Plus X-Squared" Family
  cot⁻¹ x	-1 / (1 + x²)
  sec⁻¹ x	1 / (|x|√(x² - 1))	"Absolute X Root X-Squared Minus One" Family
  cosec⁻¹ x	-1 / (|x|√(x² - 1))

  
  


• 
  Sign-Swap Pairs:
  

  Notice that for each pair (sin⁻¹/cos⁻¹, tan⁻¹/cot⁻¹, sec⁻¹/cosec⁻¹), the derivative of the 'co' function is simply the negative of its counterpart. This reduces the number of unique forms to remember to three.

  
  

JEE/Board Exam Shortcuts & Practical Tips

1. 
   Simplify Before Differentiating (Crucial for JEE):
   

   Often, expressions involving inverse trigonometric functions can be greatly simplified using trigonometric substitutions or identities before applying the differentiation rules. This is a common JEE shortcut to avoid complex chain rule applications.

   
   

   Example: Instead of directly differentiating d/dx (tan⁻¹((3x-x³)/(1-3x²))), substitute x = tanθ. The expression becomes tan⁻¹(tan3θ) = 3θ = 3tan⁻¹x. Now, d/dx(3tan⁻¹x) = 3/(1+x²), which is much simpler.

   
   


2. 
   Master the Chain Rule:
   

   Most problems will involve compositions of functions (e.g., sin(ax+b), tan⁻¹(f(x))). Ensure you can quickly apply the chain rule: d/dx [f(g(x))] = f'(g(x)) * g'(x).

   
   


3. 
   Domain Awareness (Especially for sec⁻¹x, cosec⁻¹x):
   

   Remember the absolute value in the denominators of sec⁻¹x and cosec⁻¹x derivatives: |x|√(x² - 1). This arises from the domain restrictions and definitions of these functions.

   
   

By internalizing these mnemonics and practical shortcuts, you can approach differentiation problems with greater confidence and efficiency.

Welcome to the 'Intuitive Understanding' section for the differentiation of trigonometric and inverse trigonometric functions. Our goal here is not just to memorise formulas, but to grasp the 'why' behind them, leveraging graphical insights and conceptual connections. This approach enhances retention and problem-solving skills, especially crucial for JEE Main and CBSE Board Exams.

Differentiation of Trigonometric Functions

Understanding the derivatives of basic trigonometric functions can be simplified by visualizing their graphs and the slopes at various points:

• 
  Derivative of sin(x):
  
  
  
  • Consider the graph of y = sin(x).
  
  
  • At x = 0, the sine curve is rising with a slope of 1.
  
  
  • At x = π/2, the curve momentarily flattens out, so its slope is 0.
  
  
  • At x = π, the curve is falling with a slope of -1.
  
  
  • At x = 3π/2, the curve again flattens, so its slope is 0.
  
  
  • If you plot these slopes (1, 0, -1, 0), you'll notice they perfectly trace out the graph of y = cos(x).
  
  
  • Hence, d/dx (sin(x)) = cos(x).
  
  
  
  


• 
  Derivative of cos(x):
  
  
  
  • Consider the graph of y = cos(x).
  
  
  • At x = 0, the cosine curve is momentarily flat, so its slope is 0.
  
  
  • At x = π/2, the curve is falling with a slope of -1.
  
  
  • At x = π, the curve is momentarily flat, so its slope is 0.
  
  
  • At x = 3π/2, the curve is rising with a slope of 1.
  
  
  • Plotting these slopes (0, -1, 0, 1) perfectly traces out the graph of y = -sin(x).
  
  
  • Hence, d/dx (cos(x)) = -sin(x). The negative sign intuitively arises because the cosine curve starts decreasing immediately after x=0.
  
  
  
  


• 
  Derivative of tan(x):
  
  
  
  • The graph of y = tan(x) is always increasing within its domain (between vertical asymptotes). This means its derivative must always be positive.
  
  
  • At x = 0, the slope of tan(x) is 1. The derivative sec²(x) also gives sec²(0) = 1.
  
  
  • As x approaches π/2 (or other odd multiples of π/2), tan(x) becomes very steep, tending towards infinity. Similarly, sec²(x) also tends towards infinity near these points.
  
  
  • Hence, d/dx (tan(x)) = sec²(x). The square ensures the derivative is always positive, matching the always-increasing nature of tan(x).
  
  
  
  

Differentiation of Inverse Trigonometric Functions

The key intuition for inverse function derivatives comes from the general rule: if y = f⁻¹(x), then dy/dx = 1 / (df/dy). This means the slope of an inverse function at a point is the reciprocal of the slope of the original function at the corresponding point.

• 
  Derivative of sin⁻¹(x):
  
  
  
  • Let y = sin⁻¹(x). This implies x = sin(y).
  
  
  • Differentiating x = sin(y) with respect to y, we get dx/dy = cos(y).
  
  
  • Using the inverse rule, dy/dx = 1 / (dx/dy) = 1 / cos(y).
  
  
  • To express this in terms of x, recall the identity cos(y) = ± &sqrt;(1 - sin²(y)). Since y = sin⁻¹(x) is defined such that y ∈ [-π/2, π/2], cos(y) ≥ 0. So, cos(y) = &sqrt;(1 - sin²(y)).
  
  
  • Substituting sin(y) = x, we get cos(y) = &sqrt;(1 - x²).
  
  
  • Therefore, d/dx (sin⁻¹(x)) = 1 / &sqrt;(1 - x²).
  
  
  • JEE Relevance: This derivation method is critical for understanding and quickly re-deriving formulas if you forget them under exam pressure.
  
  
  
  


• 
  Derivative of cos⁻¹(x):
  
  
  
  • Similarly, let y = cos⁻¹(x), so x = cos(y).
  
  
  • Differentiating x = cos(y) with respect to y, we get dx/dy = -sin(y).
  
  
  • Using the inverse rule, dy/dx = 1 / (dx/dy) = 1 / (-sin(y)).
  
  
  • For y = cos⁻¹(x), y ∈ [0, π], so sin(y) ≥ 0. Hence, sin(y) = &sqrt;(1 - cos²(y)) = &sqrt;(1 - x²).
  
  
  • Therefore, d/dx (cos⁻¹(x)) = -1 / &sqrt;(1 - x²). The negative sign intuitively arises from the decreasing nature of the cosine function.
  
  
  
  


• 
  Derivative of tan⁻¹(x):
  
  
  
  • Let y = tan⁻¹(x), so x = tan(y).
  
  
  • Differentiating x = tan(y) with respect to y, we get dx/dy = sec²(y).
  
  
  • Using the inverse rule, dy/dx = 1 / (dx/dy) = 1 / sec²(y).
  
  
  • Recall the identity sec²(y) = 1 + tan²(y).
  
  
  • Substituting tan(y) = x, we get sec²(y) = 1 + x².
  
  
  • Therefore, d/dx (tan⁻¹(x)) = 1 / (1 + x²).
  
  
  • The graph of tan⁻¹(x) starts relatively flat, becomes steepest at x=0, and then flattens out again. The derivative 1/(1+x²) perfectly reflects this behavior: its maximum value is 1 at x=0, and it decreases as |x| increases, always remaining positive.
  
  
  
  

By understanding these graphical and conceptual connections, you can build a strong foundation for handling differentiation problems involving these functions. This intuitive grasp will make the formulas less abstract and more logical.

The concepts of differentiation, especially involving trigonometric and inverse trigonometric functions, are not merely theoretical exercises. They form the bedrock for understanding and modeling a vast array of real-world phenomena characterized by oscillatory behavior, periodicity, and angular relationships. Calculating the rate of change of quantities that vary sinusoidally or depend on angles is crucial in numerous scientific and engineering disciplines.

Here are some key real-world applications:

• 
  Physics - Simple Harmonic Motion (SHM) and Wave Phenomena:
  
  
  
  • When a particle undergoes SHM, its position, velocity, and acceleration are described by trigonometric functions (e.g., $x(t) = A sin(omega t + phi)$). Differentiating these functions allows us to determine the instantaneous velocity ($v = dx/dt$) and acceleration ($a = dv/dt = d^2x/dt^2$) of the oscillating particle.
  
  
  • Similarly, in wave mechanics (sound waves, light waves, water waves), quantities like displacement, pressure, or electric field intensity vary sinusoidally. Differentiation helps in understanding wave propagation speeds and energy transport.
  
  
  
  


• 
  Electrical Engineering - AC Circuits:
  
  
  
  • Alternating Current (AC) and Voltage are typically sinusoidal functions of time (e.g., $I(t) = I_0 sin(omega t)$, $V(t) = V_0 sin(omega t + phi)$).
  
  
  • Differentiation is essential for analyzing circuit components like inductors and capacitors. For instance, the voltage across an inductor is given by $V_L = L frac{dI}{dt}$. If the current is sinusoidal, differentiation gives us the phase relationship and magnitude of the voltage.
  
  
  
  


• 
  Mechanical Engineering - Vibrations and Rotational Dynamics:
  
  
  
  • Analyzing the vibrations of structures (bridges, buildings, machinery) often involves trigonometric functions. Differentiation helps engineers predict resonant frequencies and design systems that can withstand dynamic loads.
  
  
  • In rotational motion, angular position, velocity, and acceleration are related through differentiation. Understanding how angles change with respect to time (angular velocity) or how angular velocity changes (angular acceleration) is fundamental.
  
  
  
  


• 
  Optics and Astronomy - Angular Rates:
  
  
  
  • Consider a lighthouse beam rotating, or tracking a satellite in orbit. The rate at which an angle changes (angular velocity) is crucial. Problems often involve an observer and a moving object, where the angle of elevation or depression is an inverse trigonometric function of the object's position. Differentiating these inverse trigonometric functions helps determine the angular speed of the observer's line of sight.
  
  
  • For example, if a camera is tracking a rocket launched vertically, its angle of elevation ($ heta$) is related to the rocket's height ($h$) and horizontal distance ($d$) by $ heta = an^{-1}(h/d)$. Differentiating with respect to time ($t$) gives $d heta/dt$, the rate at which the camera must rotate.
  
  
  
  


• 
  Computer Graphics and Animation:
  
  
  
  • Smooth transitions, rotations, and realistic motion in computer graphics often rely on trigonometric functions to model periodic or angular movements. Differentiation can be used to control the speed and acceleration of these animations, ensuring natural visual effects.
  
  
  
  

JEE Main Perspective: While direct 'real-world application' problems are less common in JEE Main, understanding these connections reinforces the importance of the concepts. It helps in developing a deeper intuition for 'rate of change' in dynamic systems, which is invaluable for problem-solving in physics and advanced engineering mathematics.

These applications demonstrate that differentiation of trigonometric and inverse trigonometric functions is a powerful tool for analyzing dynamic systems, predicting behavior, and designing solutions in a wide range of fields. Keep building your foundational understanding – it's the key to unlocking these real-world insights!

Understanding and remembering the differentiation rules for trigonometric and inverse trigonometric functions can be simplified by recognizing patterns and drawing simple analogies. These aren't always real-world comparisons but rather conceptual bridges that make the rules more intuitive and memorable.

1. The Cyclic "Dance" of Sine and Cosine Derivatives

The derivatives of sine and cosine functions exhibit a beautiful, cyclical pattern:

• Derivative of sin(x) is cos(x).


• Derivative of cos(x) is -sin(x).


• Derivative of -sin(x) is -cos(x).


• Derivative of -cos(x) is sin(x).

Analogy: Imagine a four-stage "derivative cycle" or "dance". You start with sine, take a step (differentiate) to cosine, another step to negative sine, then negative cosine, and finally back to sine. This continuous loop helps in quickly recalling the sequence and signs. For JEE Main, this pattern recognition is crucial for speed.

2. The "Co-Function" Negative Sign

Observe the derivatives of the 'co-functions' (cosine, cotangent, cosecant):

• d/dx (cos(x)) = -sin(x)


• d/dx (cot(x)) = -csc²(x)


• d/dx (csc(x)) = -csc(x)cot(x)

Analogy: Think of the "co" prefix as a signal for a negative sign in their derivatives. This simple mnemonic (memory aid) can prevent common sign errors, especially under exam pressure. While not a deep mathematical analogy, it's a very practical one for recall.

3. Inverse Function Derivative as a "Reciprocal Rate"

The general rule for differentiating an inverse function `y = f⁻¹(x)` is `dy/dx = 1 / (df/dy)`, where `x = f(y)`. This rule is the foundation for all inverse trigonometric derivatives.

Analogy: If `y = f(x)` tells you "how much `y` changes for a unit change in `x` (i.e., `dy/dx`)", then its inverse `x = f⁻¹(y)` (or `x = f(y)`) tells you "how much `x` changes for a unit change in `y` (i.e., `dx/dy`)". These two rates are reciprocally related. It's like measuring speed in "kilometers per hour" versus "hours per kilometer" – they convey the same information but are inverted. For example, knowing `dx/dy = cos(y)` for `x = sin(y)` immediately tells you `dy/dx = 1/cos(y)`, which then needs to be converted back to `x` using identities like `cos(y) = √(1-sin²(y)) = √(1-x²)`. This conceptual analogy helps understand the structure of inverse trigonometric derivatives rather than just memorizing them.

4. Chain Rule: The "Nested Doll" or "Onion Layer" Analogy

When dealing with composite functions like `sin(g(x))` or `tan⁻¹(h(x))`, the Chain Rule is indispensable. It states `d/dx [f(g(x))] = f'(g(x)) * g'(x)`.

Analogy: Differentiating a composite function is like peeling an onion or opening nested dolls. You differentiate the "outermost layer" (the main function `f`) first, treating the "inner layer" (`g(x)`) as a single variable. Then, you multiply this by the derivative of the "inner layer" (`g'(x)`). This process continues for any number of nested functions. This analogy is critical for both CBSE and JEE Main, as most practical differentiation problems involve the chain rule.

By using these analogies, you can not only memorize the differentiation rules but also gain a deeper, more intuitive understanding, which is invaluable for solving complex problems efficiently.

Prerequisites for Differentiation of Trigonometric and Inverse Trigonometric Functions

Before diving into the differentiation of trigonometric and inverse trigonometric functions, a solid foundation in the following concepts is absolutely essential. Mastery of these prerequisites will significantly ease the learning curve and prevent common errors.

1. Fundamental Differentiation Rules

A thorough understanding of basic differentiation principles is paramount. These rules will be applied constantly, often in conjunction with trigonometric and inverse trigonometric derivatives.

• Derivatives of Basic Functions: Know the derivatives of functions like $x^n$, constants, $e^x$, and $ln x$.


• Algebra of Derivatives: Master the rules for sums, differences, products, and quotients of functions:
  
  
  
  • Sum/Difference Rule: $frac{d}{dx}[f(x) pm g(x)] = frac{d}{dx}[f(x)] pm frac{d}{dx}[g(x)]$
  
  
  • Product Rule: $frac{d}{dx}[f(x) cdot g(x)] = f(x) cdot g'(x) + g(x) cdot f'(x)$
  
  
  • Quotient Rule: $frac{d}{dx}left[frac{f(x)}{g(x)}
    ight] = frac{g(x) cdot f'(x) - f(x) cdot g'(x)}{[g(x)]^2}$
  
  
  
  


• Chain Rule: This is arguably the most critical rule for differentiating composite functions. Many trigonometric and inverse trigonometric expressions are composite, e.g., $sin(2x)$, $ an^{-1}(x^2)$.
  
  
  
  • If $y = f(g(x))$, then $frac{dy}{dx} = f'(g(x)) cdot g'(x)$.
  
  
  
  

2. Trigonometric Functions and Identities

Strong recall of trigonometric concepts is non-negotiable for both CBSE and JEE Main. Simplification using identities often precedes or follows differentiation.

• Basic Trigonometric Ratios: Definition of $sin x$, $cos x$, $ an x$, $cot x$, $sec x$, $csc x$ in terms of sides of a right-angled triangle and on the unit circle.


• Fundamental Identities:
  
  
  
  • $sin^2 x + cos^2 x = 1$
  
  
  • $1 + an^2 x = sec^2 x$
  
  
  • $1 + cot^2 x = csc^2 x$
  
  
  
  


• Compound Angle Formulas:
  
  
  
  • $sin(A pm B) = sin A cos B pm cos A sin B$
  
  
  • $cos(A pm B) = cos A cos B mp sin A sin B$
  
  
  
  


• Double and Half-Angle Formulas: For example, $sin 2x = 2 sin x cos x$, $cos 2x = cos^2 x - sin^2 x = 2 cos^2 x - 1 = 1 - 2 sin^2 x$. These are frequently used for simplification.


• Domain and Range: Knowledge of the domain and range of all six trigonometric functions.

3. Inverse Trigonometric Functions and Properties

Understanding the nature of inverse trigonometric functions is key to differentiating them accurately.

• Definition of Inverse Trigonometric Functions: Understanding that they are inverses of restricted trigonometric functions to ensure they are one-to-one.


• Domain, Range, and Principal Value Branch: This is critical for inverse trigonometric functions. For example, $sin^{-1}x$ has domain $[-1, 1]$ and principal value branch $[-pi/2, pi/2]$. Misunderstanding these can lead to incorrect simplifications or solutions.


• Fundamental Properties:
  
  
  
  • $sin^{-1}x + cos^{-1}x = pi/2$
  
  
  • $ an^{-1}x + cot^{-1}x = pi/2$
  
  
  • $sec^{-1}x + csc^{-1}x = pi/2$
  
  
  • Relations like $sin^{-1}x = cos^{-1}sqrt{1-x^2}$ for $x in [0,1]$ or conversions between different inverse trig functions using right-angled triangles.
  
  
  
  


• Graphing: A basic idea of their graphs helps in visualizing their behavior and restrictions.

JEE Main Specific Tip: For JEE, be prepared to use complex trigonometric identities and inverse trigonometric properties to simplify expressions before differentiation. This often reduces a seemingly complex problem to a much simpler one.

By mastering these foundational concepts, you will be well-equipped to tackle the differentiation of trigonometric and inverse trigonometric functions efficiently and accurately.

The topic of differentiating trigonometric and inverse trigonometric functions is a cornerstone of calculus. To truly master it and excel in both board exams and JEE Main, it's crucial to understand its connections to other related mathematical concepts. These connections often serve as prerequisites, simplify complex problems, or represent immediate applications of the learned differentiation rules.

Here are the key related topics you should be proficient in:

• 
  Basic Differentiation Rules (Chain Rule, Product Rule, Quotient Rule):
  
  
  
  • Why related: Most problems involving trigonometric and inverse trigonometric functions are composite functions (e.g., sin(2x), tan⁻¹(x²)). The Chain Rule is indispensable for differentiating such functions. The Product and Quotient Rules are equally important when these functions are part of more complex expressions (e.g., x sin(x), (cos x) / x).
  
  
  • JEE Focus: JEE problems frequently combine these rules, requiring a seamless application of multiple rules in a single differentiation step.
  
  
  
  


• 
  Trigonometric Identities and Transformations:
  
  
  
  • Why related: Before differentiating, judicious use of trigonometric identities (e.g., double angle, half angle, sum/difference formulas) can often simplify a complex trigonometric expression into a much simpler form that is easier to differentiate. This is a common strategy to avoid lengthy differentiation.
  
  
  • JEE Focus: Many JEE questions on differentiation of trigonometric functions test your ability to simplify the expression using identities *before* applying differentiation, often turning a seemingly complex problem into a straightforward one. For instance, differentiating tan⁻¹((3x - x³) / (1 - 3x²)) becomes trivial if you recognize it as 3tan⁻¹(x).
  
  
  
  


• 
  Inverse Trigonometric Functions (Properties, Domain, Range):
  
  
  
  • Why related: A deep understanding of the properties, domain, range, and principal value branches of inverse trigonometric functions is essential. Misunderstanding these can lead to incorrect derivatives, especially when simplifying expressions like sin⁻¹(sin x) or cos⁻¹(cos x).
  
  
  • Common Mistake: Assuming sin⁻¹(sin x) = x without considering the domain. For example, d/dx (sin⁻¹(sin x)) is 1 only for x ∈ (-π/2, π/2). Beyond this interval, it changes.
  
  
  
  


• 
  Implicit Differentiation:
  
  
  
  • Why related: Sometimes, trigonometric or inverse trigonometric functions are part of implicitly defined relations (e.g., sin(xy) = x + y). In such cases, implicit differentiation is necessary to find dy/dx. Moreover, the derivatives of inverse trigonometric functions can be derived using implicit differentiation.
  
  
  
  


• 
  Logarithmic Differentiation:
  
  
  
  • Why related: For functions involving products, quotients, or powers where the base and exponent are functions of x, especially involving trigonometric functions (e.g., (sin x)^{tan x}), logarithmic differentiation simplifies the process by converting products/quotients into sums/differences, which are easier to differentiate.
  
  
  
  


• 
  Higher Order Derivatives:
  
  
  
  • Why related: Once you can differentiate trigonometric and inverse trigonometric functions, finding their second or higher-order derivatives (d²y/dx², d³y/dx³) is a direct extension. This is particularly common for functions like sin x, cos x, e^x, where a pattern emerges.
  
  
  
  


• 
  Applications of Derivatives:
  
  
  
  • Why related: The derivatives you learn are directly applied in various contexts, such as finding the equation of tangents and normals to curves involving trig functions, determining maxima and minima of such functions, and calculating rates of change.
  
  
  
  

When differentiating trigonometric and inverse trigonometric functions, students often fall into specific traps that can lead to incorrect answers. Being aware of these common pitfalls is crucial for securing marks in both board and competitive exams like JEE Main.

Common Exam Traps in Differentiation

• 
  Forgetting the Chain Rule: This is arguably the most frequent mistake. Students correctly differentiate the outer function but forget to multiply by the derivative of the inner function.
  
  
  
  • Trap: Differentiating sin(ax+b) as cos(ax+b) instead of a cos(ax+b).
  
  
  • Trap: Differentiating tan⁻¹(f(x)) as 1/(1 + (f(x))²) instead of f'(x) / (1 + (f(x))²).
  
  
  
  


• 
  Sign Errors in Trigonometric Derivatives: A classic oversight is getting the signs wrong for derivatives of co-functions.
  
  
  
  • Trap: Using d/dx(cos x) = sin x (incorrect) instead of d/dx(cos x) = -sin x (correct).
  
  
  • Trap: Using d/dx(cosec x) = cosec x cot x (incorrect) instead of d/dx(cosec x) = -cosec x cot x (correct).
  
  
  • Trap: Using d/dx(cot x) = sec² x (incorrect) instead of d/dx(cot x) = -cosec² x (correct).
  
  
  
  


• 
  Confusing Inverse Trigonometric Derivatives: Similar derivatives can easily be swapped, especially those that differ only by a sign.
  
  
  
  • Trap: Mixing up d/dx(sin⁻¹x) = 1/√(1-x²) with d/dx(cos⁻¹x) = -1/√(1-x²).
  
  
  • Trap: Mixing up d/dx(tan⁻¹x) = 1/(1+x²) with d/dx(cot⁻¹x) = -1/(1+x²).
  
  
  • Trap: Mixing up d/dx(sec⁻¹x) = 1/(|x|√(x²-1)) with d/dx(cosec⁻¹x) = -1/(|x|√(x²-1)).
  
  
  
  


• 
  Neglecting Pre-differentiation Simplification (JEE Specific): For inverse trigonometric functions, complex expressions can often be simplified using trigonometric substitutions or identities *before* differentiation. Failing to do so makes the differentiation process unnecessarily complicated and prone to errors.
  
  
  
  • Trap: Directly differentiating y = tan⁻¹((3x - x³)/(1 - 3x²)) instead of substituting x = tan θ to get y = tan⁻¹(tan 3θ) = 3θ = 3 tan⁻¹x, making differentiation trivial.
  
  
  
  


• 
  Ignoring Domain/Range for Inverse Trig Simplifications (JEE Specific): When simplifying expressions like sin⁻¹(sin x) or tan⁻¹(tan x), it is NOT always equal to x. The result depends on the interval of x, corresponding to the principal value branch of the inverse function.
  
  
  
  • Trap: Assuming d/dx(sin⁻¹(sin x)) = 1 for all x. This is only true when x ∈ [-π/2, π/2]. Outside this range, sin⁻¹(sin x) will be a different linear function (e.g., π - x or x - π), and its derivative will be -1 or 1 accordingly.
  
  
  • JEE Callout: For JEE, always be mindful of the range of the principal value branch when simplifying inverse trigonometric expressions. This often involves drawing graphs or checking intervals.
  
  
  
  


• 
  Misapplying Product/Quotient Rules with Trig Functions: While not unique to trig functions, the combination often leads to mistakes due to the added complexity of trig derivatives.
  
  
  
  • Trap: Differentiating x² sin x as 2x cos x (incorrect) instead of 2x sin x + x² cos x (correct, using product rule).
  
  
  
  

Key Takeaway: Always re-verify the basic derivative formulas and pay close attention to the chain rule. For inverse trigonometric functions, explore simplification opportunities using identities and be cautious about domain/range restrictions before applying differentiation rules.

Mastering the differentiation of trigonometric and inverse trigonometric functions is fundamental for success in calculus, forming a cornerstone for advanced topics like integration and differential equations. These key takeaways will consolidate the essential formulas and strategic approaches you need for both board exams and JEE Main.

Core Derivatives of Trigonometric Functions

Ensure you know these formulas by heart. They are direct applications and frequently appear.

• d/dx (sin x) = cos x




• d/dx (cos x) = -sin x




• d/dx (tan x) = sec²x




• d/dx (cot x) = -cosec²x




• d/dx (sec x) = sec x tan x




• d/dx (cosec x) = -cosec x cot x

JEE Tip: Pay close attention to the negative signs; they are common sources of error.

Core Derivatives of Inverse Trigonometric Functions

These derivatives often involve square roots and require careful handling of the argument.

• d/dx (sin⁻¹x) = 1/√(1-x²) for x ∈ (-1, 1)




• d/dx (cos⁻¹x) = -1/√(1-x²) for x ∈ (-1, 1)




• d/dx (tan⁻¹x) = 1/(1+x²) for x ∈ R




• d/dx (cot⁻¹x) = -1/(1+x²) for x ∈ R




• d/dx (sec⁻¹x) = 1/(|x|√(x²-1)) for |x| > 1




• d/dx (cosec⁻¹x) = -1/(|x|√(x²-1)) for |x| > 1

JEE Tip: Note the domain restrictions for these derivatives. While often not directly tested in simple differentiation, they are crucial for understanding function behavior.

Crucial Concepts & Strategies

Beyond memorizing formulas, applying these concepts effectively is key.

• 
  

  1. The Chain Rule is Paramount: Almost every problem involving these functions will require the Chain Rule. If y = f(g(x)), then dy/dx = f'(g(x)) * g'(x).

  
  
  
  

  • Example: d/dx (sin(x²)) = cos(x²) * d/dx(x²) = cos(x²) * 2x

  
  

  • Example: d/dx (tan⁻¹(2x)) = 1/(1+(2x)²) * d/dx(2x) = 2/(1+4x²)

  
  
  
  


• 
  

  2. Simplification through Substitution: This is a game-changer for inverse trigonometric functions, especially in JEE. Look for expressions that can be simplified using standard trigonometric identities.

  
  
  
  

  • If you see √(1-x²), consider x = sin θ or x = cos θ.

  
  

  • If you see √(1+x²), consider x = tan θ or x = cot θ.

  
  

  • If you see √(x²-1), consider x = sec θ or x = cosec θ.

  
  

  • If you see (1-x²)/(1+x²), consider x = tan θ.

  
  

  • If you see (2x)/(1+x²), consider x = tan θ.

  
  

  • Warning: When using substitution, remember to adjust the domain of the substituted variable (e.g., θ) to ensure the principal value branch of the inverse trigonometric function is maintained. This is critical for JEE problems involving piece-wise definitions.

  
  
  
  


• 
  

  3. Use Logarithmic Differentiation: For functions involving products, quotients, or powers of trigonometric/inverse trigonometric expressions (e.g., y = (sin x)^{tan x}), taking logarithms before differentiating can simplify the process.

  
  


• 
  

  4. Implicit Differentiation: When 'y' is implicitly defined in terms of 'x' within a trigonometric or inverse trigonometric equation (e.g., sin(x+y) = xy), differentiate both sides with respect to 'x', treating 'y' as a function of 'x' and applying the Chain Rule to terms involving 'y'.

  
  

CBSE vs. JEE Focus

Aspect	CBSE Board Exams	JEE Main
Direct Formulas	Often tested with simple Chain Rule applications.	Assumed knowledge; used in complex multi-step problems.
Substitution	Straightforward substitutions are common.	Critical for simplifying complex inverse trig functions; domain adjustments are often key.
Complexity	Focus on correct application of rules.	Involves nested functions, combinations with other rules (product, quotient), and implicit differentiation.

Stay sharp and practice regularly. These differentiation formulas are stepping stones to mastering calculus!

A systematic problem-solving approach is crucial when differentiating trigonometric and inverse trigonometric functions. These problems often combine basic differentiation rules with algebraic and trigonometric identities, especially in JEE Main.

General Strategy for Differentiation Problems

1. Master Basic Derivatives: Ensure you know the standard derivatives of all six trigonometric and all six inverse trigonometric functions. This is the foundation.
   
   
   
   • For example: $frac{d}{dx}(sin x) = cos x$, $frac{d}{dx}(cos x) = -sin x$, $frac{d}{dx}( an^{-1} x) = frac{1}{1+x^2}$.
   
   
   
   


2. Simplify Before Differentiating (Crucial for Inverse Trig): This is arguably the most important step for inverse trigonometric functions. Many problems are designed to be simplified using trigonometric substitutions or identities *before* applying differentiation.
   
   
   
   • JEE Focus: Recognizing forms like $ an^{-1}left(frac{2x}{1-x^2}
     ight)$, $sin^{-1}left(frac{2x}{1+x^2}
     ight)$, $cos^{-1}left(frac{1-x^2}{1+x^2}
     ight)$, which often simplify to $2 an^{-1}x$ (with domain restrictions), can save immense computational effort. Similarly, forms like $sin^{-1}(2xsqrt{1-x^2})$ can simplify to $2sin^{-1}x$.
   
   
   • Use appropriate substitutions (e.g., $x = an heta$, $x = sin heta$) to simplify the argument of the inverse trigonometric function into a single trigonometric function, which then cancels out the inverse function.
   
   
   
   


3. Identify the Primary Rule: Determine which differentiation rule applies:
   
   
   
   • Chain Rule: Almost universally applied for composite functions, e.g., $y = sin(f(x))$, $y = an^{-1}(g(x))$. Differentiate the outer function, then multiply by the derivative of the inner function.
   
   
   • Product Rule: For functions of the form $u(x)v(x)$, e.g., $x sin x$, $e^x an^{-1} x$.
   
   
   • Quotient Rule: For functions of the form $frac{u(x)}{v(x)}$, e.g., $frac{sin x}{x}$, $frac{ an^{-1} x}{x^2+1}$.
   
   
   • Implicit Differentiation: If $y$ is not explicitly defined as $f(x)$ but is part of an equation (e.g., $sin(x+y) = xy$), differentiate both sides with respect to $x$, remembering to apply the chain rule to terms involving $y$ (i.e., $frac{d}{dx}(f(y)) = f'(y) frac{dy}{dx}$).
   
   
   
   


4. Execute Differentiation Step-by-Step:
   
   
   
   • Break down complex functions into smaller, manageable parts.
   
   
   • Apply the rules carefully, one layer at a time.
   
   
   • Be meticulous with signs and algebraic manipulations.
   
   
   
   


5. Simplify the Result: After differentiating, simplify the expression algebraically as much as possible. This often involves combining terms, factoring, or using identities.

Example Walkthrough (JEE Specific):

Problem: Differentiate $y = an^{-1}left(frac{3x - x^3}{1 - 3x^2}
ight)$ with respect to $x$.

1. Analyze for Simplification: The expression $frac{3x - x^3}{1 - 3x^2}$ immediately reminds us of the $ an(3 heta)$ formula.
   
   Let $x = an heta$. Then $ heta = an^{-1}x$.
   
   Substituting, we get:
   $y = an^{-1}left(frac{3 an heta - an^3 heta}{1 - 3 an^2 heta}
   ight)$
   $y = an^{-1}( an(3 heta))$
   


2. Apply Simplification (with condition):
   
   For $y = an^{-1}( an(3 heta))$ to simplify directly to $3 heta$, we need $3 heta in left(-frac{pi}{2}, frac{pi}{2}
   ight)$, which means $ heta in left(-frac{pi}{6}, frac{pi}{6}
   ight)$.
   
   If $x = an heta$, then $ anleft(-frac{pi}{6}
   ight) < x < anleft(frac{pi}{6}
   ight)$, i.e., $-frac{1}{sqrt{3}} < x < frac{1}{sqrt{3}}$.
   
   Under this condition, $y = 3 heta = 3 an^{-1}x$.
   


3. Differentiate the Simplified Form:
   
   Now, differentiate $y = 3 an^{-1}x$ with respect to $x$:
   $frac{dy}{dx} = 3 cdot frac{d}{dx}( an^{-1}x)$
   $frac{dy}{dx} = 3 cdot frac{1}{1+x^2}$
   
   So, $frac{dy}{dx} = frac{3}{1+x^2}$ for $-frac{1}{sqrt{3}} < x < frac{1}{sqrt{3}}$.
   


4. JEE Note: If the problem does not specify the domain of $x$, a complete solution might need to consider other intervals where $y$ would be a piecewise function (e.g., $3 heta pm pi$, $3 heta pm 2pi$, etc.). However, for typical differentiation problems, the primary simplified interval is usually the expected answer unless otherwise stated. For CBSE, such domain complexities are often ignored or implicit.

Welcome to the 'JEE Focus Areas' for Differentiation of Trigonometric and Inverse Trigonometric Functions! This section will equip you with the essential knowledge and strategies to tackle JEE Main problems effectively.

JEE Focus Areas: Differentiation of Trigonometric & Inverse Trigonometric Functions

Mastering the derivatives of trigonometric and inverse trigonometric functions is fundamental for JEE Main. While direct application of formulas is important, JEE problems often involve a blend of these derivatives with the chain rule, product/quotient rule, and crucial pre-differentiation simplification techniques, especially for inverse trigonometric functions.

1. Core Derivatives to Memorize:

These are your foundational tools. Ensure you know them by heart, as any error here cascades into incorrect solutions.

• Trigonometric Functions:
  
  
  
  • $frac{d}{dx}(sin x) = cos x$
  
  
  • $frac{d}{dx}(cos x) = -sin x$
  
  
  • $frac{d}{dx}( an x) = sec^2 x$
  
  
  • $frac{d}{dx}(cot x) = -csc^2 x$
  
  
  • $frac{d}{dx}(sec x) = sec x an x$
  
  
  • $frac{d}{dx}(csc x) = -csc x cot x$
  
  
  
  


• Inverse Trigonometric Functions:
  
  
  
  • $frac{d}{dx}(sin^{-1} x) = frac{1}{sqrt{1-x^2}}$
  
  
  • $frac{d}{dx}(cos^{-1} x) = -frac{1}{sqrt{1-x^2}}$
  
  
  • $frac{d}{dx}( an^{-1} x) = frac{1}{1+x^2}$
  
  
  • $frac{d}{dx}(cot^{-1} x) = -frac{1}{1+x^2}$
  
  
  • $frac{d}{dx}(sec^{-1} x) = frac{1}{|x|sqrt{x^2-1}}$
  
  
  • $frac{d}{dx}(csc^{-1} x) = -frac{1}{|x|sqrt{x^2-1}}$
  
  
  
  

2. The All-Important Chain Rule:

JEE problems rarely involve simple $x$. You'll typically encounter functions like $sin(f(x))$, $ an^{-1}(g(x))$, etc. The Chain Rule is indispensable here. For example:

• $frac{d}{dx}(sin(ax+b)) = cos(ax+b) cdot a$


• $frac{d}{dx}( an^{-1}(x^2)) = frac{1}{1+(x^2)^2} cdot 2x = frac{2x}{1+x^4}$

3. Simplification Before Differentiation (Inverse Trigonometric Functions):

This is arguably the most critical JEE strategy for inverse trigonometric functions. Many problems are designed to be complex if differentiated directly but become trivial after simplification using trigonometric identities and properties of inverse trigonometric functions.

• Common Substitutions:
  
  
  
  • If $sqrt{a^2-x^2}$, substitute $x = asin heta$ or $x = acos heta$.
  
  
  • If $sqrt{a^2+x^2}$, substitute $x = a an heta$ or $x = acot heta$.
  
  
  • If $sqrt{x^2-a^2}$, substitute $x = asec heta$ or $x = acsc heta$.
  
  
  • For expressions like $frac{2x}{1+x^2}$, $frac{1-x^2}{1+x^2}$, $frac{3x-x^3}{1-3x^2}$, try $x = an heta$.
  
  
  
  


• Key Inverse Trig Properties: Remember formulas like $2 an^{-1}x = sin^{-1}left(frac{2x}{1+x^2}
  ight) = cos^{-1}left(frac{1-x^2}{1+x^2}
  ight) = an^{-1}left(frac{2x}{1-x^2}
  ight)$ (with domain restrictions). These directly simplify the function before you differentiate.

Example: Differentiate $y = sin^{-1}left(frac{2x}{1+x^2}
ight)$ with respect to $x$.

JEE Approach: Instead of differentiating directly, which would be complex:

1. Let $x = an heta implies heta = an^{-1}x$.


2. Substitute into the expression: $y = sin^{-1}left(frac{2 an heta}{1+ an^2 heta}
   ight)$.


3. Recognize the identity: $frac{2 an heta}{1+ an^2 heta} = sin(2 heta)$.


4. So, $y = sin^{-1}(sin(2 heta))$.


5. Assuming appropriate domain (usually implied for JEE differentiation questions, i.e., $-frac{pi}{2} le 2 heta le frac{pi}{2}$), $y = 2 heta$.


6. Substitute back $ heta = an^{-1}x$: $y = 2 an^{-1}x$.


7. Now, differentiate: $frac{dy}{dx} = frac{d}{dx}(2 an^{-1}x) = 2 cdot frac{1}{1+x^2} = frac{2}{1+x^2}$.

Direct differentiation would have been far more cumbersome.

4. CBSE vs. JEE Main Perspective:

• CBSE Board Exams: Often focus on direct application of differentiation rules, chain rule, and sometimes simple simplification for inverse functions.


• JEE Main: Emphasizes multi-concept problems. Expect differentiation combined with limits, properties of inverse trigonometric functions, and complex chain rule applications. The ability to simplify expressions before differentiating is a key differentiator in JEE.

5. Other Important Problem Types:

• Implicit Differentiation: Equations involving $y$ implicitly (e.g., $sin(x+y) = xy$) where trigonometric/inverse trigonometric functions are involved.


• Parametric Differentiation: When $x$ and $y$ are functions of a parameter (e.g., $x = f( heta), y = g( heta)$).


• Logarithmic Differentiation: For functions of the form $(f(x))^{g(x)}$ or complex products/quotients where trigonometric functions are present.

Key Takeaway: For JEE, always look for opportunities to simplify the expression, especially for inverse trigonometric functions, before applying differentiation rules. This will save significant time and prevent calculation errors.

Keep practicing and mastering these techniques for a strong score!