Find dy dx , y = ^ - 1 ( 1 - x^2 1 + x^2 ),0 < x < 1 - Vidyadip

Question

Find $\frac{{dy}}{{dx}},$ $y = {\sin ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right),0 < x < 1$

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Answer

Given: $y = {\sin ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right),0 < x < 1$
Putting $x = \tan \theta $
$y = {\sin ^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)$
$= {\sin ^{ - 1}}\left( {\cos 2\theta } \right)$
$= {\sin ^{ - 1}}\sin \left( {\frac{\pi }{2} - 2\theta } \right) = \frac{\pi }{2} - 2\theta$
$= \frac{\pi }{2} - 2{\tan ^{ - 1}}x$
$\therefore \frac{{dy}}{{dx}} = 0 - 2.\frac{1}{{1 + {x^2}}} = \frac{{ - 2}}{{1 + {x^2}}}$

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