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CBSE Board Examination 2024 · Section A · 1 mark

Q13: Derivative of sin⁡(x2)\sin(x^2) at x=πx = \sqrt{\pi}

Board 2024Ch 5 · Continuity and Differentiability
The question

The derivative of sin⁡(x2)\sin(x^2) w.r.t. x, at x=πx = \sqrt{\pi} is :

(A) 11

(B) −1-1

(C) −2π-2\sqrt{\pi}

(D) 2π2\sqrt{\pi}

Exam answer

By the chain rule,

ddxsin⁡(x2)=cos⁡(x2)⋅ddx(x2)=2xcos⁡(x2)\begin{aligned} &\frac{d}{dx}\sin(x^2) \\ &= \cos(x^2)\cdot\frac{d}{dx}(x^2) \\ &= 2x\cos(x^2) \end{aligned}
Why?

sin⁡(x2)\sin(x^2) is a function of a function: the outer function is sin⁡u\sin u and the inner one is u=x2u = x^2⁠. The chain rule says

ddxsin⁡u=cos⁡u⋅dudx\frac{d}{dx}\sin u = \cos u\cdot\frac{du}{dx}

Here u=x2u = x^2 gives du/dx=2xdu/dx = 2x⁠.

Watch out: the answer is not just cos⁡(x2)\cos(x^2)⁠. The factor 2x2x from the inner function is easy to forget.

At x=πx = \sqrt{\pi}⁠, x2=πx^2 = \pi⁠:

2πcos⁡π=−2π2\sqrt{\pi}\cos\pi = -2\sqrt{\pi}
Why?

(π)2=π(\sqrt{\pi})^2 = \pi⁠, and cos⁡π=−1\cos\pi = -1⁠. So

2π×(−1)=−2π2\sqrt{\pi}\times(-1) = -2\sqrt{\pi}

Answer: (C) −2π-2\sqrt{\pi}

Where this comes from

Chapter 5, Continuity and Differentiability: Derivative by the chain rule.

Question type: The chain rule

How to spot it: click to see

Type 4. The chain rule: a function inside a function

Spot it: something other than a plain xx inside sin⁡\sin⁠, cos⁡\cos⁠, log⁡\log⁠, ee or a power, like sin⁡(x2+5)\sin(x^2 + 5) or ecos⁡xe^{\cos x}⁠.

Here: x2x^2 sits inside sin⁡\sin⁠: the outside gives cos⁡(x2)\cos(x^2)⁠, the inside gives 2x2x⁠, so the derivative is 2xcos⁡(x2)2x\cos(x^2)⁠. At x=πx = \sqrt\pi it is 2πcos⁡π=−2π2\sqrt\pi\cos\pi = -2\sqrt\pi⁠.

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