Differentiate ${\tan ^{ - 1}}\frac{{\sqrt {1 + {x^2}} - 1}}{x}$ w.r.t. ${\tan ^{ - 1}}x,$ when $x \ne 0.$

Let $u = {\tan ^{ - 1}}\frac{{\sqrt {1 + {x^2}} - 1}}{x}$ and $v = {\tan ^{ - 1}}x$

therefore,$x = \tan \theta$

$\Rightarrow$ $u = {\tan ^{ - 1}}\frac{{\sqrt {1 + {{\tan }^2}\theta } - 1}}{{\tan \theta }}$

$= {\tan ^{ - 1}}\frac{{(\sec \theta - 1)\cos \theta }}{{\sin \theta }}$

$= {\tan ^{ - 1}}\left( {\frac{{1 - \cos \theta }}{{\sin \theta }}} \right)$

$= {\tan ^{ - 1}}\left[ {\frac{{1 - 1 + 2{{\sin }^2}\theta /2}}{{2\sin \theta /2 \cdot \cos \theta /2}}} \right]$

$= {\tan ^{ - 1}}\left[ {\tan \frac{\theta }{2}} \right]$

$= \frac{\theta }{2} = \frac{1}{2}{\tan ^{ - 1}}x$

therefore,$\frac{{du}}{{dx}} = \frac{1}{2}\frac{d}{{dx}}{\tan ^{ - 1}}x = \frac{1}{2} \cdot \frac{1}{{1 + {x^2}}}$ ………(i)

and $\frac{{dv}}{{dx}} = \frac{d}{{dx}}{\tan ^{ - 1}}x = \frac{1}{{1 + {x^2}}}$ ………(ii)

therefore,$\frac{{du}}{{dv}} = \frac{{du/dx}}{{dv/dx}}$
$= \frac{{1/2\left( {1 + {x^2}} \right)}}{{1/\left( {1 + {x^2}} \right)}} = \frac{{\left( {1 + {x^2}} \right)}}{{2\left( {1 + {x^2}} \right)}} = \frac{1}{2}$

Find $\frac{{dy}}{{dx}}$ when $x$ and $y$ are connected by the relation given