Explanation

(c)
Let $y = \tan^{- 1}\left( \frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}} \right)$
Put $x = \cos{2\theta}$
$\therefore\ \ \ y = \tan^{- 1}\left( \frac{\sqrt{1 + \cos{2\theta}} - \sqrt{1 - \cos{2\theta}}}{\sqrt{1 + \cos{2\theta}} + \sqrt{1 - \cos{2\theta}}} \right)$
$= \tan^{- 1}\left( \frac{\sqrt{2}\cos{\theta - \sqrt{2\ }\sin\theta}}{\sqrt{2\ }\cos\theta + \sqrt{2}\sin\theta} \right)$
$= \tan^{- 1}{\left( \frac{1 - \tan\theta}{1 + \tan\theta} \right) = \tan^{- 1}\left\{ \tan\left( \frac{\pi}{4} - \theta \right) \right\}}$
$= \frac{\pi}{4} - \frac{1}{2}\cos^{- 1}x\$
On differentiating w.r.t. $x,\$we get
$\ \frac{dy}{dx} = 0 + \frac{1}{2}.\frac{1}{\sqrt{1 - x}} = \frac{1}{2\sqrt{1 - x^{2}}}\$