Answer
$\frac{dy}{dt}={(1+t^2)}(\frac{-2}{1+(2t)^2})+2t\cot^{-1}{2t}$
Work Step by Step
Given $y=(1+t^2)\cot^{-1}{2t}$
On differentiating both sides:
$\frac{dy}{dt}=\frac{d((1+t^2)\cot^{-1}{2t})}{dt}$
on applying multiply rulre of derivative
$\frac{dy}{dt}={(1+t^2)}\frac{d(cot^{-1}2{t})}{dt}+\cot^{-1}{2t}\times{2t})$
$\frac{dy}{dt}={(1+t^2)}(\frac{-2}{1+(2t)^2})+2t\cot^{-1}{2t}$
