Q230 / 270 · NDA 2025 (Phase 2) · Mathematics

For the following three (03) items:
Let $y = f(x) = \dfrac{x\sin^{-1} x}{\sqrt{1 - x^2}} + \ln\sqrt{1 - x^2}$.

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

  • A
    $\theta\sec\theta$
  • B
    $\theta\sec^2\theta$
  • Correct answer
    C
    $\theta\sec^3\theta$
  • D
    $2\tan\theta + \theta\sec^2\theta$

Solution

Given $y = \dfrac{x\sin^{-1} x}{\sqrt{1 - x^2}} + \ln\sqrt{1 - x^2}$

Substitute $x = \sin\theta$ into the expression for $y$:

$y = \dfrac{\sin\theta \cdot \theta}{\sqrt{1 - \sin^2\theta}} + \ln\sqrt{1 - \sin^2\theta}$

$y = \dfrac{\theta\sin\theta}{\cos\theta} + \ln(\cos\theta)$

$y = \theta\tan\theta + \ln(\cos\theta)$

Differentiating $y$ with respect to $\theta$:

$\dfrac{dy}{d\theta} = \dfrac{d}{d\theta}(\theta\tan\theta) + \dfrac{d}{d\theta}(\ln(\cos\theta))$

$\dfrac{dy}{d\theta} = \left(\tan\theta + \theta\sec^2\theta\right) + \dfrac{1}{\cos\theta}(-\sin\theta)$

$\dfrac{dy}{d\theta} = \tan\theta + \theta\sec^2\theta - \tan\theta = \theta\sec^2\theta$

Differentiating $x = \sin\theta$ with respect to $\theta$:

$\dfrac{dx}{d\theta} = \cos\theta$

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

$\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}} = \dfrac{\theta\sec^2\theta}{\cos\theta}$

$\dfrac{dy}{dx} = \theta\sec^3\theta$

Hence, the correct answer is option: 'C'.

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