CBSE Sample Paper 2025-26 · Section A · 1 mark

Q8: $f'(1)$ for $f(x) = x\tan^{-1}x$

Sample 2025-26Ch 5 · Continuity and Differentiability

The question

If $f(x) = x\tan^{-1}x$ ⁠, then $f'(1)$ is equal to

(A) $\frac{\pi}{4} - \frac{1}{2}$

(B) $\frac{\pi}{4} + \frac{1}{2}$

(D) $-\frac{\pi}{4} + \frac{1}{2}$

Exam answer

$f(x) = x\tan^{-1}x$ ⁠. By the product rule,

\begin{aligned} f'(x) &= 1 \cdot \tan^{-1}x + x \cdot \frac{1}{1 + x^2} \end{aligned}

Why?

The product rule:

\frac{d}{dx}(uv) = \frac{du}{dx}\,v + u\,\frac{dv}{dx}

Here $u = x$ ⁠, so $\frac{du}{dx} = 1$ ⁠, and $v = \tan^{-1}x$ ⁠, so $\frac{dv}{dx} = \frac{1}{1 + x^2}$ (a standard derivative).

\begin{aligned} f'(1) &= \tan^{-1}1 + \frac{1}{1 + 1} \\ &= \frac{\pi}{4} + \frac{1}{2} \end{aligned}

Why?

Put $x = 1$ ⁠. The second term is $\frac{1}{1 + 1^2} = \frac{1}{2}$ ⁠.

$\tan^{-1}1 = \frac{\pi}{4}$ ⁠, since $\tan\frac{\pi}{4} = 1$ and $\frac{\pi}{4}$ lies in the principal range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ ⁠.

Both terms are positive, so the options with a minus sign, (A), (C) and (D), are ruled out.

Answer: (B) $\dfrac{\pi}{4} + \dfrac{1}{2}$

Where this comes from

Chapter 5, Continuity and Differentiability: Derivatives: product rule with tan⁻¹x.

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Continuity and Differentiability in past papers: question 19 of 35 · see them all

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Q8: $f'(1)$ for $f(x) = x\tan^{-1}x$