Continuity and Differentiability — Class 12 Maths Solution

$y = {\tan ^{ - 1}}\left( {\cfrac{{3x - {x^3}}}{{1 - 3{x^2}}}} \right)$
Putting $x = \tan \theta ,$ we get

$y = {\tan ^{ - 1}}\left( {\cfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}} \right)$

$\Rightarrow$ $y = {\tan ^{ - 1}}(\tan 3\theta )$
$\Rightarrow$ $y = 3\theta \Rightarrow y = 3{\tan ^{ - 1}}x$

$\Rightarrow$ $\cfrac{{dy}}{{dx}} = \cfrac{3}{{1 + {x^2}}}$