STD 12 - 5. continuity and differentiation — JEE कक्षा 11 साइन्स — Question

$x = \tan \theta $ रखने पर, तब

$y = {\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + {{\tan }^2}\theta } - 1}}{{\tan \theta }}} \right)$

==> $y = {\tan ^{ - 1}}\left( {\frac{{\sec \theta - 1}}{{\tan \theta }}} \right) = {\tan ^{ - 1}}\left( {\frac{{1 - \cos \theta }}{{\sin \theta }}} \right)$

==> $y = {\tan ^{ - 1}}\left( {\frac{{2{{\sin }^2}\frac{\theta }{2}}}{{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}} \right) = {\tan ^{ - 1}}\tan \frac{\theta }{2}$

$y = \frac{\theta }{2} = \frac{1}{2}{\tan ^{ - 1}}x$,   $(\because \theta ={{\tan }^{-1}}x)$

अत: $\frac{{dy}}{{dx}} = \frac{1}{{2(1 + {x^2})}}$.