Answer
$\frac{{dy}}{{dx}} = {\tan ^{ - 1}}x$
Work Step by Step
$$\eqalign{
& y = x{\tan ^{ - 1}}x - \frac{1}{2}\ln \left( {1 + {x^2}} \right) \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \underbrace {\frac{d}{{dx}}\left[ {x{{\tan }^{ - 1}}x} \right]}_{{\text{product rule}}} - \frac{d}{{dx}}\left[ {\frac{1}{2}\ln \left( {1 + {x^2}} \right)} \right] \cr
& \frac{{dy}}{{dx}} = x\frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}x} \right] + {\tan ^{ - 1}}x\frac{d}{{dx}}\left[ x \right] - \frac{1}{2}\frac{d}{{dx}}\left[ {\ln \left( {1 + {x^2}} \right)} \right] \cr
& {\text{Compute derivatives}} \cr
& \frac{{dy}}{{dx}} = x\left( {\frac{1}{{1 + {x^2}}}} \right) + {\tan ^{ - 1}}x\left( 1 \right) - \frac{1}{2}\left( {\frac{{2x}}{{1 + {x^2}}}} \right) \cr
& {\text{Multiply and simplify}} \cr
& \frac{{dy}}{{dx}} = \frac{x}{{1 + {x^2}}} + {\tan ^{ - 1}}x - \frac{x}{{1 + {x^2}}} \cr
& \frac{{dy}}{{dx}} = {\tan ^{ - 1}}x \cr} $$
