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NDA 2 2025 Maths, question 80

Topic: Differential calculus

Passage · Q78 to Q80

Let y=f(x)=xsin⁡−1x1−x2+ln⁡1−x2y=f(x)=\frac{x\sin ^{-1}x}{\sqrt{1-x^{2}}}+\ln \sqrt{1-x^{2}}.

Q80

If x=sin⁡θx=\sin \theta, then what is dydx\frac{dy}{dx} equal to?

  1. (a) θsec⁡θ\theta\sec \theta
  2. (b) θsec⁡2θ\theta\sec ^{2}\theta
  3. (c) θsec⁡3θ\theta\sec ^{3}\thetaCorrect answer
  4. (d) 2tan⁡θ+θsec⁡2θ2\tan \theta+\theta\sec ^{2}\theta

Solution.

Answer: (c) θsec⁡3θ\theta\sec ^{3}\theta

  1. With x=sin⁡θx = \sin\theta: sin⁡−1x=θ\sin^{-1}x = \theta and 1−x2=cos⁡θ\sqrt{1 - x^{2}} = \cos\theta.
  2. Differentiating, dydx=sin⁡−1x(1−x2)3/2=θcos⁡3θ=θsec⁡3θ\frac{dy}{dx} = \frac{\sin^{-1}x}{(1 - x^{2})^{3/2}} = \frac{\theta}{\cos^{3}\theta} = \theta\sec^{3}\theta.

Trap: This uses the principal branch, −π/2<θ<π/2-\pi/2 < \theta < \pi/2, where cos⁡θ>0\cos\theta > 0.