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

> [!TLDR] > Using the Chain Rule, we differentiate the given function with respect to $\tan^{-1} x$ by first simplifying the expression to $\frac{1}{2} \tan^{-1} x$, leading to the derivative $\frac{1}{2}$. The core idea here is the **Chain Rule** for parametric differentiation. When we need the derivative of one function $u$ with respect to another function $v$ (both of $x$), we compute: $$ \frac{du}{dv} = \frac{du/dx}{dv/dx} $$ This is just the chain rule in disguise — we differentiate both with respect to $x$ and then divide. The trick is often to simplify $u$ first, so the differentiation becomes clean. Let’s set: $$ u = \tan^{-1}\left(\frac{\sqrt{1 + x^2} - 1}{x}\right), \quad v = \tan^{-1} x $$ We want $\frac{du}{dv}$. 1. **Simplify $u$ using a trigonometric substitution.** The expression inside the arctan looks like it comes from a tangent half-angle or a double-angle identity. Let $x = \tan \theta$, so $\theta = \tan^{-1} x$. Then $\sqrt{1 + x^2} = \sqrt{1 + \tan^2 \theta} = \sec \theta$ (taking the positive root since $x$ can be any real, but we’ll handle sign later). So: $$ \frac{\sqrt{1 + x^2} - 1}{x} = \frac{\sec \theta - 1}{\tan \theta} $$ 2. **Rewrite in terms of sine and cosine.** $\sec \theta = \frac{1}{\cos \theta}$, $\tan \theta = \frac{\sin \theta}{\cos \theta}$. Then: $$ \frac{\sec \theta - 1}{\tan \theta} = \frac{\frac{1}{\cos \theta} - 1}{\frac{\sin \theta}{\cos \theta}} = \frac{\frac{1 - \cos \theta}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}} = \frac{1 - \cos \theta}{\sin \theta} $$ 3. **Use the half-angle identity.** Recall: $1 - \cos \theta = 2 \sin^2(\theta/2)$ and $\sin \theta = 2 \sin(\theta/2) \cos(\theta/2)$. So: $$ \frac{1 - \cos \theta}{\sin \theta} = \frac{2 \sin^2(\theta/2)}{2 \sin(\theta/2) \cos(\theta/2)} = \frac{\sin(\theta/2)}{\cos(\theta/2)} = \tan\left(\frac{\theta}{2}\right) $$ Therefore: $$ u = \tan^{-1}\left( \tan\left(\frac{\theta}{2}\right) \right) $$ 4.

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

NCERT Exemplar · Q53

\tan^{-1}\left(\dfrac{\sqrt{1 + x^2} - 1}{x}\right)

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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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Using the Chain Rule, we differentiate the given function with respect to $\tan^{-1} x$ by first simplifying the expression to $\frac{1}{2} \tan^{-1} x$ , leading to the derivative $\frac{1}{2}$ .

The core idea here is the Chain Rule for parametric differentiation. When we need the derivative of one function $u$ with respect to another function $v$ (both of $x$ ), we compute:

\frac{du}{dv} = \frac{du/dx}{dv/dx}

This is just the chain rule in disguise — we differentiate both with respect to $x$ and then divide. The trick is often to simplify $u$ first, so the differentiation becomes clean.

Let’s set:

u = \tan^{-1}\left(\frac{\sqrt{1 + x^2} - 1}{x}\right), \quad v = \tan^{-1} x

We want $\frac{du}{dv}$ .

Simplify $u$ using a trigonometric substitution. The expression inside the arctan looks like it comes from a tangent half-angle or a double-angle identity. Let $x = \tan \theta$ , so $\theta = \tan^{-1} x$ . Then $\sqrt{1 + x^2} = \sqrt{1 + \tan^2 \theta} = \sec \theta$ (taking the positive root since $x$ can be any real, but we’ll handle sign later). So:

\frac{\sqrt{1 + x^2} - 1}{x} = \frac{\sec \theta - 1}{\tan \theta}

Rewrite in terms of sine and cosine. $\sec \theta = \frac{1}{\cos \theta}$ , $\tan \theta = \frac{\sin \theta}{\cos \theta}$ . Then:

\frac{\sec \theta - 1}{\tan \theta} = \frac{\frac{1}{\cos \theta} - 1}{\frac{\sin \theta}{\cos \theta}} = \frac{\frac{1 - \cos \theta}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}} = \frac{1 - \cos \theta}{\sin \theta}

Use the half-angle identity. Recall: $1 - \cos \theta = 2 \sin^2(\theta/2)$ and $\sin \theta = 2 \sin(\theta/2) \cos(\theta/2)$ . So:

\frac{1 - \cos \theta}{\sin \theta} = \frac{2 \sin^2(\theta/2)}{2 \sin(\theta/2) \cos(\theta/2)} = \frac{\sin(\theta/2)}{\cos(\theta/2)} = \tan\left(\frac{\theta}{2}\right)

Therefore:

u = \tan^{-1}\left( \tan\left(\frac{\theta}{2}\right) \right)

Handle the principal value carefully. …

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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 13 on this concept.

KEAM 2026Set eng-2026-04184 marksMCQ
$y = \sin\left(\tan^{-1}\left(\dfrac{1}{\sqrt{x^2 - 1}}\right)\right),\ x > 1$
›Reveal solutionSolution

Simplify the inverse trig: the angle whose tangent is $\dfrac{1}{\sqrt{x^2-1}}$ has $\sin=\dfrac1x$ , so $y=\dfrac1x$ and its derivative is $-\dfrac{1}{x^2}$ .

Let $\theta=\tan^{-1}\!\left(\dfrac{1}{\sqrt{x^2-1}}\right)$ , so $\tan\theta=\dfrac{1}{\sqrt{x^2-1}}$ with opposite $=1$ and adjacent $=\sqrt{x^2-1}$ . The hypotenuse is $\sqrt{1+(x^2-1)}=\sqrt{x^2}=x$ (since $x>1$ ) …
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KEAM 2025Set eng-2025-04234 marksMCQ
$y = \tan^{-1}(x^2 - x)$
›Reveal solutionSolution
d/dx tan^{-1}(u) = u'/(1+u^2) with u=x^2-x gives (2x-1)/(1+(x^2-x)^2).

Concept and Intuition

The derivative of arctan(u) is u'/(1+u^2). Here u = x^2 - x so u' = 2x - 1.

Step-by-Step Solution

Let u = x^2 - x, so u' = 2x - 1.

dy/dx = u'/(1+u^2) = (2x-1)/(1+(x^2-x)^2).

Common Mistakes …
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KEAM 2025Set eng-2025-04284 marksMCQ
$x^3 = \sin\theta$
›Reveal solutionSolution

Differentiate both parametric relations with respect to $\theta$ , form $\dfrac{dy}{dx}$ , and substitute $x^6=1-y^6$ .

Concept. Here $x$ and $y$ are both given as functions of a parameter $\theta$ . For parametric curves, $\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}$ .

Step 1 — differentiate w.r.t. $\theta$ .
$x^3=\sin\theta\ \Rightarrow\ 3x^2\frac{dx}{d\theta}=\cos\theta,\qquad y^3=\cos\theta\ \Rightarrow\ 3y^2\frac{dy}{d\theta}=-\sin\theta.$
Step 2 — form $\dfrac{dy}{dx}$ .
$\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{-\sin\theta/(3y^2)}{\cos\theta/(3x^2)}=-\frac{x^2\sin\theta}{y^2\cos\theta}.$
Since $\sin\theta=x^3$ and $\cos\theta=y^3$ , …
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KEAM 2026Set eng-2026-04184 marksMCQ
$s = \sqrt{t + 1},\ x = \log s$
›Reveal solutionSolution

Substituting back, $y=6\log\sqrt{t+1}+3=3\log(t+1)+3$ , whose $t$ -derivative is $\dfrac{3}{t+1}$ .

With $s=\sqrt{t+1}$ and $x=\log s$ , we have $x=\log\sqrt{t+1}=\tfrac12\log(t+1)$ . Then

$y=6x+3=6\cdot\tfrac12\log(t+1)+3=3\log(t+1)+3.$ …
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KEAM 2026Set eng-2026-04204 marksMCQ
$x = \sec\theta - \cos\theta$
›Reveal solutionSolution

The key identities $x^2+4=(\sec\theta+\cos\theta)^2$ and $y^2+4=(\sec^{10}\theta+\cos^{10}\theta)^2$ turn $(dy/dx)^2$ into a clean ratio.

Since $x=\sec\theta-\cos\theta$ , $x^2+4=\sec^2\theta+\cos^2\theta+2=(\sec\theta+\cos\theta)^2$ .

Since $y=\sec^{10}\theta-\cos^{10}\theta$ , $y^2+4=\sec^{20}\theta+\cos^{20}\theta+2=(\sec^{10}\theta+\cos^{10}\theta)^2$ . …
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KEAM 2024Set eng-2024-06074 marksMCQ
$y=\log_e\left(\dfrac{1+2x^2}{1-3x^2}\right)$
›Reveal solutionSolution

Split the log, differentiate each term, combine over the common denominator.

$y=\log(1+2x^2)-\log(1-3x^2).$
Differentiating:
$\frac{dy}{dx}=\frac{4x}{1+2x^2}-\frac{-6x}{1-3x^2}=\frac{4x}{1+2x^2}+\frac{6x}{1-3x^2}.$
Common denominator $(1+2x^2)(1-3x^2)=1-x^2-6x^4$ ; numerator: …
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KEAM 2025Set eng-2025-04264 marksMCQ
$u = \sec^{-1}(-\sec 2\theta)$
›Reveal solutionSolution

Simplify $u=\pi-2\theta$ , differentiate both $u$ and $v$ in $\theta$ , divide.

Using $\sec^{-1}(-x)=\pi-\sec^{-1}(x)$ and $\sec^{-1}(\sec 2\theta)=2\theta$ (for $2\theta$ in the principal range),
$u=\sec^{-1}(-\sec 2\theta)=\pi-2\theta\;\Rightarrow\;\frac{du}{d\theta}=-2.$
With $v=\cos\theta$ , $\dfrac{dv}{d\theta}=-\sin\theta$ . Hence …
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KEAM 2024Set eng-2024-06084 marksMCQ
$t^2+t$
›Reveal solutionSolution

Differentiate parametrically: divide $\dfrac{d(t^2+t)}{dt}$ by $\dfrac{d(t-1)}{dt}$ , then substitute $t=-2$ .

Let $u=t^2+t$ and $w=t-1$ . The derivative of $u$ with respect to $w$ is

$\frac{du}{dw}=\frac{du/dt}{dw/dt}.$ …
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KEAM 2025Set eng-2025-04284 marksMCQ
$y = \sin x \sin 2x$
›Reveal solutionSolution

Express $y$ in $t=\cos x$ : $y=2t-2t^3$ , then $\dfrac{dy}{dt}=2-6t^2=2(1-3t^2)$ .

With $t=\cos x$ and using $\sin 2x=2\sin x\cos x$ :
$y=\sin x\,\sin 2x=\sin x\,(2\sin x\cos x)=2\sin^2x\cos x.$
Since $\sin^2x=1-\cos^2x=1-t^2$ and $\cos x=t$ , …
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KEAM 2025Set eng-2025-04264 marksMCQ
$x \in \mathbb{R}$
›Reveal solutionSolution

With $f(x)=\log3-\sin x$ , $f'(x)=-\cos x$ ; by the chain rule $g'(0)=f'(f(0))f'(0)=(-\cos(\log3))(-\cos0)=\cos(\log3)$ .

Here $f(x)=\log3-\sin x$ so $f'(x)=-\cos x$ . Then $f(0)=\log3-\sin0=\log3$ and $f'(0)=-\cos0=-1$ . …
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KEAM 2025Set eng-2025-04234 marksMCQ
$h(x) = f\left(\sqrt{g(x)}\right)$
›Reveal solutionSolution
By the chain rule h'(3) = f'(3)g'(3)/(2sqrt(g(3))) = 6*3/6 = 3.

Concept and Intuition

h(x) = f(sqrt(g(x))) is a triple composition; differentiate outer-to-inner, picking up the derivative of the square root and of g.

Step-by-Step Solution

h'(x) = f'(sqrt(g(x))) * d/dx[sqrt(g(x))] = f'(sqrt(g)) * g'(x)/(2*sqrt(g(x))).

At x = 3: g(3) = 9 so sqrt(g) = 3, f'(3) = 6, g'(3) = 3. …
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KEAM 2021Set eng-2021-P2-B14 marksMCQ
$f(x) = \left(x^3 + \sin \pi x\right)^5$
›Reveal solutionSolution
$f'(1)=5(3-\pi)$ .

Concept and Intuition

Differentiate the outer fifth power via the chain rule and multiply by the derivative of the inner expression, then evaluate at $x=1$ .

Step-by-Step Solution

$f'(x)=5(x^3+\sin\pi x)^4\cdot(3x^2+\pi\cos\pi x)$ .

At $x=1$ : inner base $=1+\sin\pi=1$ , so $1^4=1$ .

Inner derivative $=3(1)+\pi\cos\pi=3-\pi$ . …
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Q.Differentiate $\tan^{-1}\left(\dfrac{\sqrt{1 + x^2} - 1}{x}\right)$ w.r.t. $\tan^{-1} x$ , when $x \ne 0$ .

Concept understanding — Chain Rule

The Chain Rule: Why It Makes Sense

The Precise Statement

A Simple Example

Why It's Called a "Chain"

Method: Differentiating One Function with Respect to Another

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.Differentiate tan⁡−1(1+x2−1x)\tan^{-1}\left(\dfrac{\sqrt{1 + x^2} - 1}{x}\right)tan−1(x1+x2​−1​) w.r.t. tan⁡−1x\tan^{-1} xtan−1x, when x≠0x \ne 0x=0.

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

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