Differentiate w.r.t. $x$: $\tan^{-1}\left(\dfrac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}}\right),\ -1 < x < 1,\ x \ne 0$.

Differentiate w.r.t. x: tan⁻¹(dfrac√(1 + x²) + √(1 – Class 12 Mathematics

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Concept understanding — Chain Rule

The Chain Rule: Why It Makes Sense

Imagine you're assembling a toy. First you put part A into part B, then you put that combined piece into part C. The final toy's position depends on how you moved A, which then affected B, which then affected C. That's exactly what the chain rule captures — how a change in the first variable ripples through a sequence of functions to affect the final output.

Let's make this concrete. Suppose you have a function $f$ that depends on $g$ , and $g$ itself depends on $x$ :

y = f(g(x))

You want to know: if $x$ changes by a tiny amount, how much does $y$ change? The answer isn't just $f'(g(x))$ — because $g(x)$ itself changes when $x$ changes. You have to multiply the two rates:

How fast does $g$ change with respect to $x$ ? That's $g'(x)$ .
How fast does $f$ change with respect to its input $g$ ? That's $f'(g(x))$ .

The total effect is the product:

\frac{dy}{dx} = f'(g(x)) \cdot g'(x)

Note

In Leibniz notation, this looks even more natural: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ , where $u = g(x)$ . The $du$ 's "cancel" like fractions — though this is just a helpful memory aid, not a rigorous proof.

The Precise Statement

Chain Rule (single variable): If $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$ , then the composite function $h(x) = f(g(x))$ is differentiable at $x$ , and

h'(x) = f'(g(x)) \cdot g'(x)

That's it. One multiplication. But the power is enormous — it lets you differentiate almost any nested function.

A Simple Example

Differentiate $h(x) = \sin(3x^2)$ .

Here $f(u) = \sin u$ and $g(x) = 3x^2$ . Then:

$f'(u) = \cos u$ , so $f'(g(x)) = \cos(3x^2)$
$g'(x) = 6x$

Multiply: $h'(x) = \cos(3x^2) \cdot 6x = 6x \cos(3x^2)$

Watch out

The most common mistake is forgetting to multiply by the inner derivative. Students often write $\frac{d}{dx} \sin(3x^2) = \cos(3x^2)$ and stop — that's wrong. The chain rule demands you also multiply by $6x$ .

Why It's Called a "Chain"

Think of a chain of links: $x \to g \to f$ . Each link has its own rate of change. To find the total rate from $x$ to $f$ , you multiply the rates of each link. If you had three functions — say $h(x) = f(g(k(x)))$ — you'd multiply three derivatives: …

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Substituting $x^2 = \cos 2\theta$ collapses the messy fraction inside $\tan^{-1}$ into $\tan\!\left(\tfrac{\pi}{4}+\theta\right)$ , so $y = \tfrac{\pi}{4} + \tfrac12\cos^{-1}(x^2)$ and the Chain Rule gives $\dfrac{dy}{dx} = -\dfrac{x}{\sqrt{1-x^4}}$ .

Attacking this directly with the quotient rule and two nested square roots would be brutal. The smart move is to simplify the argument of $\tan^{-1}$ before differentiating. Whenever you see $\sqrt{1+x^2}$ and $\sqrt{1-x^2}$ appearing together, reach for a substitution that turns both radicals into clean trig functions.

1. Pick the right substitution

Let $x^2 = \cos 2\theta$ . Because $-1 < x < 1$ with $x \ne 0$ , we have $0 < x^2 < 1$ , so $\cos 2\theta \in (0,1)$ , giving $2\theta \in (0,\pi/2)$ and therefore

\theta \in \left(0,\tfrac{\pi}{4}\right),\qquad \cos\theta > 0,\ \sin\theta > 0.

Using the half-angle identities $1+\cos 2\theta = 2\cos^2\theta$ and $1-\cos 2\theta = 2\sin^2\theta$ :

\sqrt{1+x^2} = \sqrt{2\cos^2\theta} = \sqrt{2}\,\cos\theta,\qquad \sqrt{1-x^2} = \sqrt{2\sin^2\theta} = \sqrt{2}\,\sin\theta.

The positivity of $\cos\theta$ and $\sin\theta$ on $(0,\pi/4)$ lets us drop the absolute values safely.

2. Simplify the fraction

Substitute and cancel the common $\sqrt{2}$ :

\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}} = \frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}.

Divide top and bottom by $\cos\theta$ (nonzero here):

\frac{1+\tan\theta}{1-\tan\theta}.

The tangent addition formula $\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$ with $\alpha=\tfrac{\pi}{4}$ (so $\tan\alpha=1$ ) and $\beta=\theta$ gives exactly

\frac{1+\tan\theta}{1-\tan\theta} = \tan\left(\frac{\pi}{4}+\theta\right).

3. Remove the inverse tangent — carefully

So the function is

y = \tan^{-1}\left[\tan\left(\frac{\pi}{4}+\theta\right)\right].

The identity $\tan^{-1}(\tan u)=u$ is only valid when $u \in (-\tfrac{\pi}{2},\tfrac{\pi}{2})$ , so we must check the range. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 115 on this concept.

CBSE 2020Set 65/3/11 mark
$\sin^2(\sqrt{x})$
›Reveal solutionSolution
We have a composition of three functions: squaring, sine, and square root. The chain rule peels them off one layer at a time, giving $\frac{d}{dx}\sin^2(\sqrt{x}) = \sin(\sqrt{x})\cos(\sqrt{x}) \cdot \frac{1}{\sqrt{x}}$ .

The chain rule is our tool for differentiating composite functions. When a function is built by nesting one operation inside another, we differentiate from the outside in, multiplying the derivative of each layer as we go.

Here $\sin^2(\sqrt{x})$ is really $[\sin(\sqrt{x})]^2$ , so we have three nested functions:

Outermost: squaring something

Middle: taking the sine of something

Innermost: taking the square root of $x$

The chain rule says: differentiate the outer function (leaving the inside alone), then multiply by the derivative of what's inside, and repeat until you reach $x$ .

Step-by-step differentiation:

Differentiate the outer square.

If $u = \sin(\sqrt{x})$ , then we're differentiating $u^2$ . The power rule gives $2u$ , so:

$\frac{d}{dx}[\sin(\sqrt{x})]^2 = 2\sin(\sqrt{x}) \cdot \frac{d}{dx}[\sin(\sqrt{x})]$

Differentiate the sine layer.

Now we need $\frac{d}{dx}[\sin(\sqrt{x})]$ . The derivative of $\sin(v)$ is $\cos(v)$ , where $v = \sqrt{x}$ :

$\frac{d}{dx}[\sin(\sqrt{x})] = \cos(\sqrt{x}) \cdot \frac{d}{dx}[\sqrt{x}]$

Differentiate the innermost square root.

Finally, $\frac{d}{dx}[\sqrt{x}] = \frac{d}{dx}[x^{1/2}] = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$ . …
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CBSE 2024Set 65/1/11 markMCQ
$\sin(x^2)$
›Reveal solutionSolution
The derivative of $\sin(x^2)$ is found using the Chain Rule: differentiate the outer sine, then multiply by the derivative of the inner $x^2$ . At $x = \sqrt{\pi}$ , the result is $-2\sqrt{\pi}$ , which corresponds to option (C).

The key to this problem is recognizing that $\sin(x^2)$ is a composite function. You have an outer function, $\sin(\text{something})$ , and an inner function, $x^2$ . When you need the derivative of a composition like this, the Chain Rule is your only reliable tool.

Why does the Chain Rule work? Think of it as peeling an onion: you first differentiate the outer layer (sine) while keeping the inner layer untouched, then multiply by the derivative of the inner layer. This gives the rate of change of the whole expression with respect to $x$ .

Let’s walk through it step by step.

Identify the outer and inner functions.

Here, $f(u) = \sin(u)$ where $u = x^2$ . So $f'(u) = \cos(u)$ , and $u' = 2x$ .

Apply the Chain Rule.

The derivative is:

$\frac{d}{dx} \sin(x^2) = \cos(x^2) \cdot \frac{d}{dx}(x^2) = \cos(x^2) \cdot 2x.$
So $\frac{d}{dx} \sin(x^2) = 2x \cos(x^2)$ .

Evaluate at $x = \sqrt{\pi}$ . Substitute $x = \sqrt{\pi}$ into the derivative:

$2(\sqrt{\pi}) \cos\left( (\sqrt{\pi})^2 \right) = 2\sqrt{\pi} \cos(\pi).$

Simplify $\cos(\pi)$ . From the unit circle, $\cos(\pi) = -1$ . So:
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CBSE 20191 mark
$y = \cos (\sqrt{3x})$
›Reveal solutionSolution
Use the chain rule: differentiate the outer cosine function, then multiply by the derivative of the inner function $\sqrt{3x}$ . The result is $\frac{dy}{dx} = -\frac{\sqrt{3}}{2\sqrt{x}} \sin(\sqrt{3x})$ .

The key idea here is the chain rule. Whenever you have a function of a function — like $\cos$ of something that itself depends on $x$ — you differentiate layer by layer. Think of it as peeling an onion: first the outer layer (cosine), then the next layer (the square root), and finally the innermost layer ( $3x$ ). Each derivative multiplies together.

Let’s walk through it step by step.

Identify the composition. We have $y = \cos(u)$ , where $u = \sqrt{3x}$ . So $y$ depends on $u$ , and $u$ depends on $x$ . The chain rule says:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.$

Differentiate the outer function. The derivative of $\cos(u)$ with respect to $u$ is $-\sin(u)$ . So:

$\frac{dy}{du} = -\sin(u) = -\sin(\sqrt{3x}).$

Differentiate the inner function $u = \sqrt{3x}$ . Write $\sqrt{3x}$ as $(3x)^{1/2}$ . Using the power rule and chain rule again (or directly):

$\frac{du}{dx} = \frac{1}{2}(3x)^{-1/2} \cdot 3 = \frac{3}{2\sqrt{3x}}.$
Simplify: $\frac{3}{2\sqrt{3x}} = \frac{\sqrt{3}}{2\sqrt{x}}$ , because $\frac{3}{\sqrt{3}} = \sqrt{3}$ . …
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CBSE 2026Set A1 markMCQ
$\frac{d}{dx}\sqrt{x^2+ax+1} =$
(a) $\frac{x+a}{2\sqrt{x^2+ax+1}}$
(b) $\frac{2x+a}{2\sqrt{x^2+ax+1}}$
(c) $\frac{2x+a}{\sqrt{x^2+ax+1}}$
(d) $\frac{1}{2\sqrt{x^2+ax+1}}$
›Reveal solutionSolution

$\frac{d}{dx}\sqrt{x^2+ax+1} = \frac{2x+a}{2\sqrt{x^2+ax+1}}$ .

Let $u = x^2+ax+1$ , so $\frac{du}{dx} = 2x+a$ .

Using $\frac{d}{dx}\sqrt{u} = \frac{1}{2\sqrt{u}}\cdot\frac{du}{dx}$ : …

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CBSE 2026Set A1 markMCQ
$\frac{d}{dx}(\sin x^2) =$
(a) $2x\cos x^2$
(b) $\cos x^2$
(c) $x^2\cos x^2$
(d) $x\cos x^2$
›Reveal solutionSolution

$\frac{d}{dx}\sin(x^2) = 2x\cos(x^2)$ .

Let $u = x^2$ , so $\frac{du}{dx} = 2x$ .

By the chain rule, …

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CBSE 2026Set A1 markMCQ
$\frac{d}{dx}\sqrt{\cot x} =$
(a) $\frac{1}{2\sqrt{\cot x}}$
(b) $\sqrt{\csc^2 x}$
(c) $\frac{-\csc^2 x}{2\sqrt{\cot x}}$
(d) $\frac{\csc^2 x}{2\sqrt{\cot x}}$
›Reveal solutionSolution

$\frac{d}{dx}\sqrt{\cot x} = \frac{-\csc^2 x}{2\sqrt{\cot x}}$ .

Let $u = \cot x$ , so $\frac{du}{dx} = -\csc^2 x$ .

Using $\frac{d}{dx}\sqrt{u} = \frac{1}{2\sqrt{u}}\cdot\frac{du}{dx}$ : …

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CBSE 2026Set A1 markMCQ
$\frac{d}{dx}(\cos\sqrt{x}) =$
(a) $\sin\sqrt{x}$
(b) $\frac{-\sin\sqrt{x}}{2\sqrt{x}}$
(c) $\frac{\sin\sqrt{x}}{2\sqrt{x}}$
(d) $\frac{1}{2\sqrt{x}}$
›Reveal solutionSolution

$\frac{d}{dx}\cos\sqrt{x} = \frac{-\sin\sqrt{x}}{2\sqrt{x}}$ .

Let $u = \sqrt{x}$ , so $\frac{du}{dx} = \frac{1}{2\sqrt{x}}$ .

By the chain rule, …

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CBSE 2026Set A1 markMCQ
$\frac{d}{dx}(\cos x^3) =$
(a) $-3x^2\sin x^3$
(b) $\sin x^3$
(c) $3x^2\sin x^3$
(d) $3x^2$
›Reveal solutionSolution

Chain rule on $\cos(x^3)$ gives $-3x^2\sin x^3$ .

Let the inner function be $u=x^3$ , so $y=\cos u$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\log(\cos e^x)$
(a) $-\tan e^x$
(b) $e^x\tan e^x$
(c) $-e^x\tan e^x$
(d) $\tan e^x$
›Reveal solutionSolution

Using the chain rule twice, $\dfrac{d}{dx}\log(\cos e^x)=-e^x\tan e^x$ .

Let $y=\log(\cos e^x)$ . We differentiate using the chain rule, working from the outside in.

Step 1: Differentiate $\log(u)$ where $u=\cos e^x$ :

$\frac{dy}{dx}=\frac{1}{\cos e^x}\cdot\frac{d}{dx}(\cos e^x)$

Step 2: Differentiate $\cos(v)$ where $v=e^x$ : …

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CBSE 2026Set ANNUAL1 markMCQ
$y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
(a) $\frac{1}{1+x^2}$
(b) $\frac{2}{4+x}$
(c) $\frac{2}{1+x^2}$
(d) $\frac{2}{x+x^2}$
›Reveal solutionSolution

Put $x=\tan\phi$ so the expression simplifies to $\cos^{-1}(\cos2\phi)=2\phi=2\tan^{-1}x$ , whose derivative is standard.

Let $x=\tan\phi$ . Then $\dfrac{1-x^2}{1+x^2} = \dfrac{1-\tan^2\phi}{1+\tan^2\phi} = \cos2\phi$ .

…

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CBSE 2026Set ANNUAL1 mark
$y = e^x + e^{x^2} + \ldots + e^{x^5}$
›Reveal solutionSolution

Differentiate each term $e^{x^n}$ using the chain rule: $\dfrac{d}{dx}e^{x^n}=nx^{n-1}e^{x^n}$ .

$y=e^x+e^{x^2}+e^{x^3}+e^{x^4}+e^{x^5}$

…

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CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{d}{dx}(\cos 3x) =$
(a) $\sin 3x$
(b) $-3\sin 3x$
(c) $\cos 3x$
(d) $-3\cos 3x$
›Reveal solutionSolution

Differentiate $\cos(3x)$ using the chain rule: derivative of $\cos u$ is $-\sin u$ , times $\dfrac{du}{dx}$ .

…

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Q.Differentiate w.r.t. $x$ : $\tan^{-1}\left(\dfrac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}}\right),\ -1 < x < 1,\ x \ne 0$ .

Concept understanding — Chain Rule

The Chain Rule: Why It Makes Sense

The Precise Statement

A Simple Example

Why It's Called a "Chain"

1. Pick the right substitution

2. Simplify the fraction

3. Remove the inverse tangent — carefully

Method: Trigonometric Substitution Inside Inverse Trig Functions

Steps

Common Mistakes

Q.Differentiate w.r.t. xxx: tan⁡−1(1+x2+1−x21+x2−1−x2), −1<x<1, x≠0\tan^{-1}\left(\dfrac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}}\right),\ -1 < x < 1,\ x \ne 0tan−1(1+x2​−1−x2​1+x2​+1−x2​​), −1<x<1, x=0.

1. Pick the right substitution

2. Simplify the fraction

3. Remove the inverse tangent — carefully

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Q.Differentiate w.r.t. $x$ : $\tan^{-1}\left(\dfrac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}}\right),\ -1 < x < 1,\ x \ne 0$ .

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