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NDA 2025
Exam
NDA
Year
2025
Subject
Mathematics
Chapter
Differentiation

NDA 2025 · Mathematics · Differentiation

For the following three (03) items: Let y = f(x) = x ^ -1 x 1 - x^2...

For the following three (03) items:
Let y=f(x)=xsin⁡−1x1−x2+ln⁡1−x2y = f(x) = \dfrac{x\sin^{-1} x}{\sqrt{1 - x^2}} + \ln\sqrt{1 - x^2}.

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

Options

  1. Aθsec⁡θ\theta\sec\theta
  2. Bθsec⁡2θ\theta\sec^2\theta
  3. Cθsec⁡3θ\theta\sec^3\theta
  4. D2tan⁡θ+θsec⁡2θ2\tan\theta + \theta\sec^2\theta

Correct answer

C. θsec⁡3θ\theta\sec^3\theta

Step-by-step solution

Given y=xsin⁡−1x1−x2+ln⁡1−x2y = \dfrac{x\sin^{-1} x}{\sqrt{1 - x^2}} + \ln\sqrt{1 - x^2}

Substitute x=sin⁡θx = \sin\theta into the expression for yy:

y=sin⁡θ⋅θ1−sin⁡2θ+ln⁡1−sin⁡2θy = \dfrac{\sin\theta \cdot \theta}{\sqrt{1 - \sin^2\theta}} + \ln\sqrt{1 - \sin^2\theta}

y=θsin⁡θcos⁡θ+ln⁡(cos⁡θ)y = \dfrac{\theta\sin\theta}{\cos\theta} + \ln(\cos\theta)

y=θtan⁡θ+ln⁡(cos⁡θ)y = \theta\tan\theta + \ln(\cos\theta)

Differentiating yy with respect to θ\theta:

dydθ=ddθ(θtan⁡θ)+ddθ(ln⁡(cos⁡θ))\dfrac{dy}{d\theta} = \dfrac{d}{d\theta}(\theta\tan\theta) + \dfrac{d}{d\theta}(\ln(\cos\theta))

dydθ=(tan⁡θ+θsec⁡2θ)+1cos⁡θ(−sin⁡θ)\dfrac{dy}{d\theta} = \left(\tan\theta + \theta\sec^2\theta\right) + \dfrac{1}{\cos\theta}(-\sin\theta)

dydθ=tan⁡θ+θsec⁡2θ−tan⁡θ=θsec⁡2θ\dfrac{dy}{d\theta} = \tan\theta + \theta\sec^2\theta - \tan\theta = \theta\sec^2\theta

Differentiating x=sin⁡θx = \sin\theta with respect to θ\theta:

dxdθ=cos⁡θ\dfrac{dx}{d\theta} = \cos\theta

Using the chain rule to find dydx\dfrac{dy}{dx}:

dydx=dydθdxdθ=θsec⁡2θcos⁡θ\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}} = \dfrac{\theta\sec^2\theta}{\cos\theta}

dydx=θsec⁡3θ\dfrac{dy}{dx} = \theta\sec^3\theta
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