Answer
$\frac{dy}{dt}=(\frac{t}{1+{t^2}}+\tan^{-1}{t}-\frac{1}{2t})$
Work Step by Step
Given $y=t\tan^{-1}{t}-\frac{1}{2}\ln{t}$
On differentiating both sides:
$\frac{dy}{dt}=\frac{d(t\tan^{-1}{t}-\frac{1}{2}\ln{t})}{dt}$
$\frac{dy}{dt}=(\frac{t}{1+{t^2}}+\tan^{-1}{t}-\frac{1}{2t})$
