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CBSE Sample Paper 2026-27 · Section A · 1 mark

Q7: Derivative of cos⁡−1(2x2−1)\cos^{-1}(2x^2 - 1) with respect to cos⁡−1x\cos^{-1}x

Sample 2026-27Ch 5 · Continuity and DifferentiabilityCh 2 · Inverse Trigonometric Functions
The question

The derivative of cos⁡−1(2x2−1)\cos^{-1}(2x^2 - 1) with respect to cos⁡−1(x)\cos^{-1}(x) is

(A) 22

(B) ±1\pm 1

(C) 12\frac{1}{2}

(D) −3-3

Exam answer

Let y=cos⁡−1(2x2−1)y = \cos^{-1}(2x^2 - 1) and u=cos⁡−1xu = \cos^{-1}x⁠, so x=cos⁡ux = \cos u⁠.

Why?

We need dydu\frac{dy}{du}⁠, the derivative of one function with respect to another. The quickest way is to write yy directly in terms of uu⁠, and u=cos⁡−1xu = \cos^{-1}x means cos⁡u=x\cos u = x⁠.

Take 0<x<10 < x < 1⁠, so 0<u<π20 < u < \frac{\pi}{2} and 0<2u<π0 < 2u < \pi⁠.

y=cos⁡−1(2cos⁡2u−1)=cos⁡−1(cos⁡2u)=2u\begin{aligned} y &= \cos^{-1}(2\cos^2u - 1) \\ &= \cos^{-1}(\cos 2u) = 2u \end{aligned}
Why?

Replace xx by cos⁡u\cos u⁠, then use the double angle formula

x=cos⁡u⇒x2=cos⁡2u2x2−1=2cos⁡2u−1\begin{aligned} x = \cos u &\Rightarrow x^2 = \cos^2u \\ 2x^2 - 1 &= 2\cos^2u - 1 \end{aligned} 2cos⁡2u−1=cos⁡2u2\cos^2u - 1 = \cos 2u

cos⁡−1(cos⁡θ)=θ\cos^{-1}(\cos\theta) = \theta only when θ\theta lies in [0,π][0, \pi]⁠, the principal range of cos⁡−1\cos^{-1}⁠. That is why we need 0<2u<π0 < 2u < \pi⁠.

⇒dydu=2\Rightarrow \frac{dy}{du} = 2
Why?

The derivative of 2u2u with respect to uu is 2.

Check with dydu=dy/dxdu/dx\frac{dy}{du} = \frac{dy/dx}{du/dx}⁠. For 0<x<10 < x < 1⁠,

Use ddxcos⁡−1t=−11−t2⋅dtdx\frac{d}{dx}\cos^{-1}t = \frac{-1}{\sqrt{1 - t^2}}\cdot\frac{dt}{dx} with t=2x2−1t = 2x^2 - 1⁠, dtdx=4x\frac{dt}{dx} = 4x⁠:

1−t2=1−(2x2−1)2=1−(4x4−4x2+1)=4x2−4x4=4x2(1−x2)1−t2=2x1−x2[x>0]dydx=−4x2x1−x2\begin{aligned} 1 - t^2 &= 1 - (2x^2 - 1)^2 \\ &= 1 - (4x^4 - 4x^2 + 1) \\ &= 4x^2 - 4x^4 \\ &= 4x^2(1 - x^2) \\[4pt] \sqrt{1 - t^2} &= 2x\sqrt{1 - x^2} \quad [x > 0] \\[4pt] \frac{dy}{dx} &= \frac{-4x}{2x\sqrt{1 - x^2}} \end{aligned} dydx=−21−x2dudx=−11−x2\begin{aligned} \frac{dy}{dx} &= \frac{-2}{\sqrt{1 - x^2}} \\ \frac{du}{dx} &= \frac{-1}{\sqrt{1 - x^2}} \end{aligned}

and the ratio is 2 again:

dy/dxdu/dx=−2−1=2\frac{dy/dx}{du/dx} = \frac{-2}{-1} = 2

Answer: (A) 22

The official scheme writes cos⁡−1(cos⁡2u)=2u\cos^{-1}(\cos 2u) = 2u without a condition. This step holds only for 0≤x≤10 \le x \le 1⁠; for −1<x<0-1 < x < 0 the derivative is −2-2⁠, which is not an option.

Where this comes from

Chapter 5, Continuity and Differentiability: Derivative of one function with respect to another.

Question type: Inverse trig of a fraction

How to spot it: click to see

Type 6. Inverse trig of a fraction: put x=tan⁡θx = \tan\theta first

Spot it: sin⁡−1\sin^{-1}⁠, cos⁡−1\cos^{-1} or tan⁡−1\tan^{-1} of a fraction such as 2x1+x2\frac{2x}{1 + x^2} or 1−x21+x2\frac{1 - x^2}{1 + x^2}⁠.

Here: 2x2−12x^2 - 1 is 2cos⁡2u−12\cos^2u - 1 in disguise, so put x=cos⁡ux = \cos u (here a cos, not a tan). Then y=cos⁡−1(cos⁡2u)=2uy = \cos^{-1}(\cos 2u) = 2u⁠, and as u=cos⁡−1xu = \cos^{-1}x⁠, the derivative is 2.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 2, Inverse Trigonometric Functions.

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