$y = {\sin ^{ - 1}}\left( {\cfrac{{2x}}{{1 + {x^2}}}} \right)$

$y = {\sin ^{ - 1}}\left( {\cfrac{{2x}}{{1 + {x^2}}}} \right)$

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

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

therefore, $\cfrac{{dy}}{{dx}} = \cfrac{2}{{1 + {x^2}}}$