If $f(x) = |\tan 2x|$, then find the value of $f'(x)$ at $x = \dfrac{\pi}{3}$. OR If $y = \operatorname{cosec}(\cot^{-1} x)$, then prove that $\sqrt{1 + x^2}\,\dfrac{dy}{dx} - x = 0$.

If f(x) = |tan 2x|, then find the value of f'(x) at x = (π)/(3). OR If y = operatornamecosec(cot⁻¹ x), then pr

Question

f(x) = |\tan 2x|

(OR)

y = \operatorname{cosec}(\cot^{-1} x)

CBSECBSE Class XII Board 2024Subjective· 2mImportance★★★★★

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Part (a)Concept understanding — Differentiability of Absolute Value

Differentiability of the Absolute Value Function

Start with something familiar: the absolute value of $x$ , written $|x|$ , is its distance from zero on the number line. So $|3| = 3$ , $|-5| = 5$ , and $|0| = 0$ . Graphically, it looks like a V-shape — two straight lines meeting at the origin.

Differentiability is about whether a function has a well-defined slope (derivative) at a point. For smooth curves like $x^2$ or $\sin x$ , the slope exists everywhere. But the absolute value function has a sharp corner at $x = 0$ — and that corner is the whole story.

Intuition: Why the corner matters

Walk along $y = |x|$ from left to right. Approaching $x = 0$ from the left, the slope is $-1$ (the line goes downward). Leaving $x = 0$ to the right, the slope is suddenly $+1$ (the line goes upward). At $x = 0$ , there's no single slope — it changes abruptly. That's why $|x|$ is not differentiable at $x = 0$ . Everywhere else — for $x < 0$ and $x > 0$ — the graph is a straight line with constant slope, so $|x|$ is differentiable at every point except $x = 0$ .

Note

A function must be continuous to be differentiable, but continuity alone isn't enough. The absolute value function is continuous at $x = 0$ (no break), yet fails to be differentiable there because of the sharp corner.

The precise statement

Let $f(x) = |x|$ . Then:

For $x > 0$ : $f(x) = x$ , so $f'(x) = 1$ .
For $x < 0$ : $f(x) = -x$ , so $f'(x) = -1$ .
At $x = 0$ : the derivative does not exist, because the left-hand and right-hand derivatives are different numbers.

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

Watch out

A common mistake is to think that because $|x|$ is continuous at $x=0$ , it must be differentiable there. Continuity is necessary for differentiability, but not sufficient. The absolute value function is the classic counterexample.

The bigger picture …

Part (b)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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Part (a): near $x=\tfrac\pi3$ , $\tan2x<0$ so $f=-\tan2x$ and $f'(\tfrac\pi3)=-8$ . Part (b): $y=\operatorname{cosec}(\cot^{-1}x)$ simplifies to $\sqrt{1+x^2}$ , from which $\sqrt{1+x^2}\,\dfrac{dy}{dx}-x=0$ .

Part (a)

Concept

$|u|$ has a corner only where $u=0$ . Since $\tan2x\ne0$ at $x=\tfrac\pi3$ , $f$ is differentiable there, and the local sign of $\tan2x$ fixes the sign of the derivative.

Steps

Value: $2x=\tfrac{2\pi}{3}$ , $\tan\tfrac{2\pi}{3}=-\tan\tfrac\pi3=-\sqrt3\ (\ne0).$
Sign nearby: for $x$ just below and above $\tfrac\pi3$ , $2x$ stays in the second quadrant where $\tan<0$ . So $|\tan2x|=-\tan2x$ locally.
Differentiate: $f'(x)=-2\sec^2 2x.$ …

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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 124 on this concept.

CBSE 2026Set V11 mark
$[0, 3, -1, 2, -2, 1]$
›Reveal solutionSolution

Just left of $0$ , $|x|=-x$ has slope $-1$ , so the left-hand derivative is $-1$ .

The left-hand derivative at $x=0$ is

$\lim_{h\to0^-}\frac{|0+h|-|0|}{h}=\lim_{h\to0^-}\frac{-h}{h}=-1,$ …
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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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CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{d}{dx} \tan^{-1}(x^2) =$
(a) $\dfrac{2x}{1+x^4}$
(b) $\dfrac{x}{1+x^2}$
(c) $\dfrac{x^3}{1+x^2}$
(d) None of these
›Reveal solutionSolution

Use $\dfrac{d}{dx}\tan^{-1}u = \dfrac{1}{1+u^2}\cdot\dfrac{du}{dx}$ with $u=x^2$ .

…

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

Differentiate $e^{u}$ with $u=\sin x$ : derivative is $e^u \cdot \dfrac{du}{dx}$ .

…

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Q.If f(x)=∣tan⁡2x∣f(x) = |\tan 2x|f(x)=∣tan2x∣, then find the value of f′(x)f'(x)f′(x) at x=π3x = \dfrac{\pi}{3}x=3π​.

Part (a)

Concept

Steps

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Q.If $f(x) = |\tan 2x|$ , then find the value of $f'(x)$ at $x = \dfrac{\pi}{3}$ .

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