What is the derivative of $${\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} - 1}}{x}} \right)$$     with respect to $${\tan ^{ - 1}}x\,?$$

Solution :
$$\eqalign{ & {\text{Let }}y = {\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 + {x^2}} - 1}}{x}} \right]{\text{ and }}u = {\tan ^{ - 1}}x \cr & {\text{Put }}x = \tan \,\theta \, \Rightarrow \theta = {\tan ^{ - 1}}x \cr & {\text{Then, }}y = {\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 + {{\tan }^2}\theta } - 1}}{{\tan \,\theta }}} \right] \cr & = {\tan ^{ - 1}}\left[ {\frac{{\sqrt {{{\sec }^2}\theta } - 1}}{{\tan \,\theta }}} \right] \cr & = {\tan ^{ - 1}}\left[ {\frac{{\sec \,\theta - 1}}{{\tan \,\theta }}} \right] \cr & = {\tan ^{ - 1}}\left[ {\frac{{\frac{1}{{\cos \,\theta }} - 1}}{{\frac{{\sin \,\theta }}{{\cos \,\theta }}}}} \right] \cr & = {\tan ^{ - 1}}\left[ {\frac{{1 - \cos \,\theta }}{{\sin \,\theta }}} \right] \cr & = {\tan ^{ - 1}}\left[ {\frac{{2\,{{\sin }^2}\frac{\theta }{2}}}{{2\,\sin \frac{\theta }{2}.\cos \frac{\theta }{2}}}} \right] \cr & \left( {\because \,1 - \cos \,\theta = 2\,{{\sin }^2}\frac{\theta }{2}{\text{ and }}\sin \,x = 2\,\sin \frac{x}{2}.\,\cos \frac{x}{2}} \right) \cr & = {\tan ^{ - 1}}\left[ {\tan \frac{\theta }{2}} \right] \cr & \Rightarrow \,y = \frac{\theta }{2} \cr & \Rightarrow y = \frac{{{{\tan }^{ - 1}}x}}{2}\,\,\,\,\,\,\left[ {\because \,\theta = {{\tan }^{ - 1}}x} \right] \cr & \Rightarrow y = \frac{u}{2}\,;\,\,\,\frac{{dy}}{{du}} = \frac{1}{2} \cr} $$