MCQ
${d \over {dx}}\left( {{{\tan }^{ - 1}}{{\sqrt {1 + {x^2}} - 1} \over x}} \right) = . . .$
- A${1 \over {1 + {x^2}}}$
- ✓${1 \over {2(1 + {x^2})}}$
- C${{{x^2}} \over {2\sqrt {1 + {x^2}} (\sqrt {1 + {x^2}} - 1)}}$
- D${2 \over {1 + {x^2}}}$
Put $x = \tan \theta ,$ then
$y = {\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + {{\tan }^2}\theta } - 1}}{{\tan \theta }}} \right)$
$y = {\tan ^{ - 1}}\left( {\frac{{\sec \theta - 1}}{{\tan \theta }}} \right) = {\tan ^{ - 1}}\left( {\frac{{1 - \cos \theta }}{{\sin \theta }}} \right)$
$y = {\tan ^{ - 1}}\left( {\frac{{2{{\sin }^2}\frac{\theta }{2}}}{{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}} \right) = {\tan ^{ - 1}}\tan \frac{\theta }{2}$
$y = \frac{\theta }{2} = \frac{1}{2}{\tan ^{ - 1}}x$, $(\because \theta ={{\tan }^{-1}}x)$
Hence $\frac{{dy}}{{dx}} = \frac{1}{{2(1 + {x^2})}}$.
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