MCQ
${d \over {dx}}\left( {{{\tan }^{ - 1}}{{\sqrt {1 + {x^2}} - 1} \over x}} \right)$ is equal to
  • 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
    ${2 \over {1 + {x^2}}}$

Answer

Correct option: B.
${1 \over {2(1 + {x^2})}}$
b
(b) Let $y = {\tan ^{ - 1}}\frac{{\sqrt {1 + {x^2}} - 1}}{x}$

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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