Differentiate $\sec^{-1} \left( \frac{1}{\sqrt{1-x^2}} \right)$ w.r.t. $\sin^{-1}(2x\sqrt{1-x^2})$.

$\frac{1}{2}$

The correct answer is Option (2) → $\frac{1}{2}$ ##

Let $u = \sec^{-1} \left( \frac{1}{\sqrt{1-x^2}} \right)$ and $v = \sin^{-1}(2x\sqrt{1-x^2})$

Let $x = \sin \theta$

$u = \sec^{-1} \left( \frac{1}{\sqrt{1-\sin^2 \theta}} \right)$

$ = \sec^{-1} \left( \frac{1}{\cos \theta} \right) = \theta$

$∴u = \sin^{-1} x$

$v = \sin^{-1}(2\sin \theta \sqrt{1-\sin^2 \theta})$

$ = \sin^{-1}(2 \sin \theta \cos \theta)$

$v = \sin^{-1}(\sin 2\theta) = 2\theta$

$v = 2 \sin^{-1} x$

$\frac{du}{dx} = \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}$

$\frac{dv}{dx} = \frac{d}{dx}(2 \sin^{-1} x) = \frac{2}{\sqrt{1-x^2}}$

$\frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{1}{2}$