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

$x = \sin \theta $ रखने पर,

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

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

अत:, $y = \frac{{{{\sin }^{ - 1}}x}}{2} \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {1 - {x^2}} }}$.