Continuity and Differentiability — Class 12 Maths Solution

$y = {\cos ^{ - 1}}\left( {\cfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right),$ where $0 < x < 1$

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

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

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