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

$x = \cos \theta \Rightarrow \theta = {\cos ^{ - 1}}x$ रखने पर,

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

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

==> $\frac{{dy}}{{d\theta }} = 2\sin \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right)\,.\,\cos \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right)\,\left( { - \frac{1}{2}} \right)$

==> $\frac{{dy}}{{d\theta }} = - \frac{{\sin (\pi - \theta )}}{2} = - \frac{{\sin \theta }}{2} = \frac{{ - 1}}{2}\,\sqrt {1 - {x^2}} $

==> $\frac{{dy}}{{dx}} = \frac{{dy}}{{d\theta }}\,.\,\frac{{d\theta }}{{dx}} = \frac{{ - 1}}{2}\sqrt {1 - {x^2}} \,.\,\frac{d}{{dx}}({\cos ^{ - 1}}x) = \frac{1}{2}$.