Find $\frac{dy}{dx}$, if $y = e^{\sin^2 x} \left\{ 2 \tan^{-1} \sqrt{\frac{1 - x}{1 + x}} \right\}$.

$e^{\sin^2 x} \cos^{-1} x \left[ \sin 2x - \frac{1}{\cos^{-1} x \sqrt{1 - x^2}} \right]$

The correct answer is Option (1) → $e^{\sin^2 x} \cos^{-1} x \left[ \sin 2x - \frac{1}{\cos^{-1} x \sqrt{1 - x^2}} \right]$ ##

Putting $x = \cos 2\theta$ in $\left\{ 2 \tan^{-1} \sqrt{\frac{1 - x}{1 + x}} \right\}$, we get:

$2 \tan^{-1} \sqrt{\frac{1 - \cos 2\theta}{1 + \cos 2\theta}}$

i.e., $2 \tan^{-1} \sqrt{\frac{2 \sin^2 \theta}{2 \cos^2 \theta}} = 2 \tan^{-1}(\tan \theta)$

$= 2\theta = \cos^{-1} x$

Then, $y = e^{\sin^2 x} \cos^{-1} x$

Or, $\log y = \sin^2 x + \log(\cos^{-1} x)$

Or, $\frac{1}{y} \frac{dy}{dx} = 2 \sin x \cos x + \frac{1}{\cos^{-1} x} \cdot \frac{-1}{\sqrt{1 - x^2}}$

$= \sin 2x - \frac{1}{\cos^{-1} x \sqrt{1 - x^2}}$

Or, $\frac{dy}{dx} = e^{\sin^2 x} \cos^{-1} x \left[ \sin 2x - \frac{1}{\cos^{-1} x \sqrt{1 - x^2}} \right]$