$y = {\tan ^{ - 1}}\left( {{{\tan x - 1} \over {\tan x + 1}}} \right)$ $\Rightarrow$ $- {\tan ^{ - 1}}\left( {{{1 - \tan x} \over {1 + \tan x}}} \right)$ $\Rightarrow$ $- {\tan ^{ - 1}}\left( {\tan \left( {{\pi \over 4} - x} \right)} \right)$ $\Rightarrow$ $- \left( {{\pi \over 4} - x} \right)$ $\Rightarrow$ ${{dy} \over {dx}} = 1$ Now if differentiation of ${x \over 2}$ w.r.t is ${1 \over 2}$ $\Rightarrow$ So differentiation of y w.r.t ${x \over 2}$ is $1 \over {1 \over 2}$ = 2