Continuity and Differentiability — Class 12 Maths Solution

Putting $x = \tan \theta ,$we get

$y = {\sin ^{ - 1}}\left( {\cfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right) \Rightarrow y = {\sin ^{ - 1}}(\cos 2\theta )$

$\Rightarrow$ $y = {\sin ^{ - 1}}\left\{ {\sin \left( {\cfrac{\pi }{2} - 2\theta } \right)} \right\} \Rightarrow y = \cfrac{\pi }{2} - 2\theta$

$\Rightarrow$ $y = \cfrac{\pi }{2} - 2{\tan ^{ - 1}}x \Rightarrow \cfrac{{dy}}{{dx}} = 0 - \cfrac{2}{{1 + {x^2}}} \Rightarrow \cfrac{{dy}}{{dx}} = - \cfrac{2}{{1 + {x^2}}}$