Answer
$$\frac{{dy}}{{dt}} = \frac{t}{{1 + {t^2}}} + {\tan ^{ - 1}}t - \frac{1}{{2t}}$$
Work Step by Step
$$\eqalign{
& y = t{\tan ^{ - 1}}t - \frac{1}{2}\ln t \cr
& {\text{find the derivative of }}y{\text{ with respect to }}t \cr
& \frac{{dy}}{{dt}} = \frac{{d\left( {t{{\tan }^{ - 1}}t - \frac{1}{2}\ln t} \right)}}{{dt}} \cr
& \frac{{dy}}{{dt}} = \frac{{d\left( {t{{\tan }^{ - 1}}t} \right)}}{{dt}} - \frac{{d\left( {\frac{1}{2}\ln t} \right)}}{{dt}} \cr
& \frac{{dy}}{{dt}} = \frac{{d\left( {t{{\tan }^{ - 1}}t} \right)}}{{dt}} - \frac{1}{2}\frac{{d\left( {\ln t} \right)}}{{dt}} \cr
& {\text{use the product rule}} \cr
& \frac{{dy}}{{dt}} = t\frac{{d\left( {{{\tan }^{ - 1}}t} \right)}}{{dt}} + {\tan ^{ - 1}}t\frac{{d\left( t \right)}}{{dt}} - \frac{1}{2}\frac{{d\left( {\ln t} \right)}}{{dt}} \cr
& {\text{use the formula }}\cr
&\frac{{d\left( {{{\tan }^{ - 1}}x} \right)}}{{dx}} = \frac{1}{{1 + {x^2}}}\frac{{du}}{{dx}}{\text{ and }}\cr
&\frac{{d\left( {\ln x} \right)}}{{dx}} = \frac{1}{x} \cr
& \frac{{dy}}{{dt}} = t\left( {\frac{1}{{1 + {t^2}}}} \right) + {\tan ^{ - 1}}t\left( 1 \right) - \frac{1}{2}\left( {\frac{1}{t}} \right) \cr
& {\text{simplifying, we get:}} \cr
& \frac{{dy}}{{dt}} = \frac{t}{{1 + {t^2}}} + {\tan ^{ - 1}}t - \frac{1}{{2t}} \cr} $$
