If $f(x) = \sqrt{\dfrac{\sec x - 1}{\sec x + 1}}$, find $f'\left(\dfrac{\pi}{3}\right)$. OR Find $f'(x)$ if $f(x) = (\tan x)^{\tan x}$.

> [!TLDR] > Part (a): $f'\!\left(\tfrac{\pi}{3}\right)=\dfrac{2}{3}$. Part (b): $f'(x)=(\tan x)^{\tan x}\sec^2 x(1+\ln\tan x)$. ## Part (a) **Idea.** Simplify the radical with half-angle identities *before* differentiating. 1. Convert $\sec x=\tfrac{1}{\cos x}$: $$f(x)=\sqrt{\frac{\frac1{\cos x}-1}{\frac1{\cos x}+1}}=\sqrt{\frac{1-\cos x}{1+\cos x}}.$$ 2. Use $1-\cos x=2\sin^2\tfrac{x}{2}$ and $1+\cos x=2\cos^2\tfrac{x}{2}$: $$f(x)=\sqrt{\tan^2\tfrac{x}{2}}=\Big|\tan\tfrac{x}{2}\Big|.$$ 3. At $x=\tfrac{\pi}{3}$, $\tfrac{x}{2}=\tfrac{\pi}{6}$ and $\tan\tfrac{\pi}{6}>0$, so locally $f(x)=\tan\tfrac{x}{2}$. 4. Differentiate with the chain rule: $$f'(x)=\sec^2\tfrac{x}{2}\cdot\tfrac12=\tfrac12\sec^2\tfrac{x}{2}.$$ …

If f(x) = √((sec x - 1)/(sec x + 1)), find f'((π)/(3)). — CBSE Class 12 Mathematics 2020 PYQ

Question

f(x) = \sqrt{\dfrac{\sec x - 1}{\sec x + 1}}

(OR)

f'(x)

CBSECBSE Class XII Board 2020Subjective· 2mImportance★★★★★

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Part (a)Concept understanding — Derivative Evaluation

Derivative Evaluation

To evaluate a derivative means to find $f'(a)$ — a single number that tells you how fast $f$ is changing right at $x=a$ . Think of a speedometer: it doesn't report how far you've travelled, only how fast your position is changing at this instant. That instantaneous rate of change is exactly what $f'(a)$ measures.

The geometric picture

On the curve $y=f(x)$ , pick a point $P$ and a nearby point $Q$ . The straight line through them — the secant — has slope equal to the average rate of change between $P$ and $Q$ . Now slide $Q$ toward $P$ : the secant rotates into the tangent line that just touches the curve at $P$ , and its slope is $f'(a)$ .

Note

$f'(a)$ is the slope of the tangent to $y=f(x)$ at $x=a$ — how steep the curve is right there.

The limit definition

f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

Here $h$ is a tiny step from $a$ to $a+h$ , the numerator is the matching change in height, and the ratio is a secant slope. As $h\to 0$ the secant becomes the tangent. An equivalent form is

f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}.

When this limit exists, $f$ is differentiable at $a$ (which forces continuity there).

Watch out

Continuity alone is not enough. $f(x)=|x|$ is continuous at $0$ , but its left slope $-1$ and right slope $+1$ disagree, so $f'(0)$ does not exist — a corner has no single tangent.

A worked evaluation

For $f(x)=x^2$ at $x=3$ :

f'(3) = \lim_{h \to 0} \frac{(3+h)^2 - 9}{h} = \lim_{h \to 0} (6 + h) = 6.

So the tangent at $x=3$ has slope $6$ .

From a number to a function …

Part (b)Concept understanding — Derivative Evaluation

Derivative Evaluation

The geometric picture

Note

$f'(a)$ is the slope of the tangent to $y=f(x)$ at $x=a$ — how steep the curve is right there.

The limit definition

f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

Here $h$ is a tiny step from $a$ to $a+h$ , the numerator is the matching change in height, and the ratio is a secant slope. As $h\to 0$ the secant becomes the tangent. An equivalent form is

f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}.

When this limit exists, $f$ is differentiable at $a$ (which forces continuity there).

Watch out

Continuity alone is not enough. $f(x)=|x|$ is continuous at $0$ , but its left slope $-1$ and right slope $+1$ disagree, so $f'(0)$ does not exist — a corner has no single tangent.

A worked evaluation

For $f(x)=x^2$ at $x=3$ :

f'(3) = \lim_{h \to 0} \frac{(3+h)^2 - 9}{h} = \lim_{h \to 0} (6 + h) = 6.

So the tangent at $x=3$ has slope $6$ .

From a number to a function …

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Part (a): $f'\!\left(\tfrac{\pi}{3}\right)=\dfrac{2}{3}$ . Part (b): $f'(x)=(\tan x)^{\tan x}\sec^2 x(1+\ln\tan x)$ .

Part (a)

Idea. Simplify the radical with half-angle identities before differentiating.

Convert $\sec x=\tfrac{1}{\cos x}$ :

f(x)=\sqrt{\frac{\frac1{\cos x}-1}{\frac1{\cos x}+1}}=\sqrt{\frac{1-\cos x}{1+\cos x}}.

Use $1-\cos x=2\sin^2\tfrac{x}{2}$ and $1+\cos x=2\cos^2\tfrac{x}{2}$ :

f(x)=\sqrt{\tan^2\tfrac{x}{2}}=\Big|\tan\tfrac{x}{2}\Big|.

At $x=\tfrac{\pi}{3}$ , $\tfrac{x}{2}=\tfrac{\pi}{6}$ and $\tan\tfrac{\pi}{6}>0$ , so locally $f(x)=\tan\tfrac{x}{2}$ .
Differentiate with the chain rule: $f'(x)=\sec^2\tfrac{x}{2}\cdot\tfrac12=\tfrac12\sec^2\tfrac{x}{2}.$ …

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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 103 on this concept.

CBSE 2026Set 65/3/11 markMCQ
$\sin^{-1} x = y$
›Reveal solutionSolution
The derivative of $\sin^{-1} x$ is found by implicit differentiation of $x = \sin y$ , giving $\frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\sqrt{1-x^2}}$ , which matches option (B) $\cos y$ only if we interpret it as $\sec y$ — but careful: the correct form is $\frac{1}{\sqrt{1-x^2}}$ , and among the given choices, (B) $\cos y$ is actually $\frac{1}{\cos y}$ ? No — let's check: $\cos y = \sqrt{1-x^2}$ , so $\frac{1}{\cos y} = \sec y$ , which is option (D). The final answer is (D) $\sec y$ .

The core idea: when you have an inverse trigonometric function, the easiest way to differentiate it is to rewrite it as a direct trigonometric equation and then use implicit differentiation. This avoids memorising a dozen formulas and builds from what you already know — the derivative of $\sin$ and the chain rule.

Let $\sin^{-1} x = y$ . This means $x = \sin y$ , and importantly, $y$ is restricted to $[-\pi/2, \pi/2]$ so that $\cos y \ge 0$ .

Start with the relation:

$x = \sin y$

Differentiate both sides with respect to $x$ . Remember $y$ is a function of $x$ , so we use the chain rule on the right:

$\frac{d}{dx}(x) = \frac{d}{dx}(\sin y)$ $1 = \cos y \cdot \frac{dy}{dx}$

Solve for $\frac{dy}{dx}$ :

$\frac{dy}{dx} = \frac{1}{\cos y}$

Now, $\cos y$ can be expressed in terms of $x$ . Since $\sin y = x$ , we use the identity $\sin^2 y + \cos^2 y = 1$ :

$\cos^2 y = 1 - \sin^2 y = 1 - x^2$ $\cos y = \sqrt{1 - x^2} \quad (\text{positive because } y \in [-\pi/2, \pi/2])$

Therefore: $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$ …
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CBSE 2026Set V11 markMCQ
$x - y = \pi$
(a) $\pi$
(b) $-\pi$
(c) $1$
(d) $-1$
›Reveal solutionSolution

Differentiating the constant-difference relation gives $\dfrac{dy}{dx}=1$ ; answer (c).

Differentiate $x-y=\pi$ (a constant) with respect to $x$ : …
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CBSE 2026Set A1 markMCQ
$\frac{d}{dx}(\log x^n) =$
(a) $\frac{1}{x^n}$
(b) $n$
(c) $\frac{1}{x}$
(d) $\frac{n}{x}$
›Reveal solutionSolution

$\frac{d}{dx}(\log x^n) = \frac{n}{x}$ .

First simplify with the log power rule:

$\log x^n = n\log x.$

Differentiate: …

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

$\frac{d}{dx}\left(e^{x-a}\right) = e^{x-a}$ .

Here $a$ is a constant. Let $u = x-a$ , so $\frac{du}{dx} = 1$ .

By the chain rule, …

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CBSE 2026Set A1 markMCQ
$\frac{d}{dx}(\tan^{-1}\sqrt{x} + \cot^{-1}\sqrt{x}) =$
(a) $\frac{\pi}{2}$
(b) $0$
(c) $1$
(d) $\pi$
›Reveal solutionSolution

The derivative is $0$ because the sum is a constant.

For any real argument $t$ , $\tan^{-1}t + \cot^{-1}t = \frac{\pi}{2}$ . Taking $t=\sqrt{x}$ ,

$\tan^{-1}\sqrt{x} + \cot^{-1}\sqrt{x} = \frac{\pi}{2}.$

…

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

$\frac{d}{dx}(2\tan^{-1}x) = \frac{2}{1+x^2}$ .

Using the standard derivative $\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}$ and the constant multiple rule: …

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CBSE 2026Set A1 markMCQ
$\frac{d}{dx}\left\{\lim_{x\to 0}\frac{x^5-a^5}{x-a}\right\} =$
(a) $a$
(b) $0$
(c) $5a^4$
(d) $5$
›Reveal solutionSolution

The inner limit is a constant, so the derivative is $0$ .

Evaluate the limit first (it does not depend on $x$ after the limit is taken):

$\lim_{x\to a}\frac{x^5-a^5}{x-a} = 5a^4$ …

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

$\frac{d}{dx}\cot^{-1}x = \frac{-1}{1+x^2}$ .

This is a standard result. From $\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}$ , differentiating gives …

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CBSE 2026Set A1 markMCQ
$\frac{d}{dx}\sin^{-1}(3x - 4x^3) =$
(a) $\frac{3}{\sqrt{1-x^2}}$
(b) $\frac{-3}{\sqrt{1-x^2}}$
(c) $\frac{1}{\sqrt{1-x^2}}$
(d) $\frac{-1}{\sqrt{1-x^2}}$
›Reveal solutionSolution

$\frac{d}{dx}\sin^{-1}(3x-4x^3) = \frac{3}{\sqrt{1-x^2}}$ .

Use the identity (for the principal branch, $-\frac{1}{2}\le x\le\frac{1}{2}$ ):

$\sin^{-1}(3x - 4x^3) = 3\sin^{-1}x,$

which follows from $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ with $x=\sin\theta$ .

Differentiate: …

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CBSE 2026Set A1 markMCQ
$\frac{d}{dx}(\sin 3x \cdot \cos 5x) =$
(a) $4\cos 8x$
(b) $4\cos 8x - \cos 2x$
(c) $\cos 2x$
(d) $\cos 2x - 4\cos 8x$
›Reveal solutionSolution

Convert the product to a sum, then differentiate: $\frac{d}{dx}(\sin 3x\cos 5x)=4\cos 8x-\cos 2x$ .

Use the product-to-sum identity $\sin A\cos B=\tfrac12[\sin(A+B)+\sin(A-B)]$ .

$\sin 3x\cos 5x=\tfrac12[\sin 8x+\sin(-2x)]=\tfrac12[\sin 8x-\sin 2x]$ .

Differentiate term by term:

…

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CBSE 2026Set ANNUAL1 mark
$y = \frac{e^x}{\sin x}$
›Reveal solutionSolution

Apply the quotient rule to $y=e^x/\sin x$ .

$\dfrac{dy}{dx} = \dfrac{e^x\sin x - e^x\cos x}{\sin^2 x} = \dfrac{e^x(\sin x-\cos x)}{\sin^2x}$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{d}{dx}(a^x) =$
(a) $a^x \log a$
(b) $\log a$
(c) $a^x$
(d) None of these
›Reveal solutionSolution

$\dfrac{d}{dx}a^x = a^x \ln a$ , using $a^x = e^{x\ln a}$ .

…

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Q.If f(x)=sec⁡x−1sec⁡x+1f(x) = \sqrt{\dfrac{\sec x - 1}{\sec x + 1}}f(x)=secx+1secx−1​​, find f′(π3)f'\left(\dfrac{\pi}{3}\right)f′(3π​).

Part (a)

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Q.If $f(x) = \sqrt{\dfrac{\sec x - 1}{\sec x + 1}}$ , find $f'\left(\dfrac{\pi}{3}\right)$ .

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