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Given: {tex}y = {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right){/tex}
To simplify the given Inverse Trigonometric function,we put, {tex}x = \tan \theta{/tex}
{tex}\Rightarrow y = {\sin ^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right) = {\sin ^{ - 1}}\left( {\sin 2\theta } \right) = 2\theta{/tex}
{tex}\Rightarrow y = 2{\tan ^{ - 1}}x{/tex}
{tex}\Rightarrow \frac{{dy}}{{dx}} = 2.\frac{1}{{1 + {x^2}}} = \frac{2}{{1 + {x^2}}}{/tex}
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