Answer
$1-2t\coth^{-1}t$
Work Step by Step
We are given: $y=(1-t^2) \coth^{-1} t$
Recall the formula: $\dfrac{d (\coth^{-1} x)}{dx}=\dfrac{1}{1-x^2}$
We need to use the product rule to get the differentiation:
$\dfrac{dy}{dt}=(1-t^2) \dfrac{1}{1-t^2} +\coth^{-1}t (-2t)=1-2t\coth^{-1}t$
