Explanation

(a)
In the given equation put $x = \tan{\theta,\ }$we get
$y = \tan^{- 1}\left( \frac{\sqrt{1 + \tan^{2}\theta} - 1}{\tan\theta} \right)$
$= \tan^{- 1}{\left( \frac{\sec\theta - 1}{\tan\theta} \right) = \tan^{- 1}{\tan{\frac{\theta}{2} = \frac{\theta}{2}}}}$
$\Rightarrow y = \frac{1}{2}\tan^{- 1}x$
$\Rightarrow y^{'} = \frac{1}{2(1 + x^{2})}\ \ \ \ \ \Rightarrow y^{'}(0) = \frac{1}{2}$