Let f = ${\tan ^{ - 1}}\left( {{{\sqrt {1 + {x^2}} - 1} \over x}} \right)$ Put x = tan $\theta$ $\Rightarrow$ $\theta$ = tan –1 x f = ${\tan ^{ - 1}}\left( {{{\sec \theta - 1} \over {\tan \theta }}} \right)$ $\Rightarrow$ f = ${\tan ^{ - 1}}\left( {{{1 - \cos \theta } \over {\sin \theta }}} \right)$ = ${\theta \over 2}$ $\Rightarrow$ f = ${{{{\tan }^{ - 1}}x} \over 2}$ $\therefore$ ${{df} \over {dx}}$ = ${1 \over {2\left( {1 + {x^2}} \right)}}$ ....(1) Let g = ${\tan ^{ - 1}}\left( {{{2x\sqrt {1 - {x^2}} } \over {1 - 2{x^2}}}} \right)$ Put x = sin $\theta$ $\Rightarrow$ $\theta$ = sin –1 x $\Rightarrow$ g = ${\tan ^{ - 1}}\left( {{{2\sin \theta \cos \theta } \over {1 - 2{{\sin }^2}\theta }}} \right)$ $\Rightarrow$ g = tan –1 (tan 2$\theta$) = 2$\theta$ $\Rightarrow$ g = 2sin -1 x $\Rightarrow$ ${{dg} \over {dx}}$ = ${2 \over {\sqrt {1 - {x^2}} }}$ ...(2) Using (i) and (ii), $\therefore$ ${{df} \over {dg}}$ = ${1 \over {2\left( {1 + {x^2}} \right)}}{{\sqrt {1 - {x^2}} } \over 2}$ At x = ${1 \over 2}$, ${\left( {{{df} \over {dg}}} \right)_{x = {1 \over 2}}}$ = ${{\sqrt 3 } \over {10}}$