Explanation

(c)
Let $y = \tan^{- 1}\left( \frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}} \right)$
Put $x = \cos{2\theta \Rightarrow \theta = \frac{1}{2}\cos^{- 1}x}$
$\therefore y = \tan^{- 1}\left( \frac{\sqrt{1 + \cos{2\theta}} - \sqrt{1 - \cos{2\theta}}}{\sqrt{1 + \cos{2\theta}} + \sqrt{1 - \cos{2\theta}}} \right)$
$\Rightarrow y = \tan^{- 1}\left( \frac{\sqrt{2\cos^{2}\theta} - \sqrt{2\sin^{2}\theta}}{\sqrt{2\cos^{2}\theta} + \sqrt{2\sin^{2}\theta}} \right)$
$\Rightarrow y = \tan^{- 1}\left( \frac{\cos{\theta - \sin\theta}}{\cos{\theta + \sin\theta}} \right)$
$\Rightarrow y = \tan^{- 1}\left( \frac{1 - \tan\theta}{1 + \tan\theta} \right)$
$\Rightarrow y = \tan^{- 1}\left( \tan\left( \frac{\pi}{4} - \theta \right) \right)$
$\Rightarrow y = \frac{\pi}{4} - \theta$
$\Rightarrow y = \frac{\pi}{4} - \frac{1}{2}\cos^{- 1}x$
On differentiating w.r.t. $x$, we get
$\frac{dy}{dx} = - \frac{1}{2}\left( \frac{- 1}{\sqrt{1 - x^{2}}} \right) = \frac{1}{2\sqrt{1 - x^{2}}}$