Answer
$-sech^2 \dfrac{1}{t}+2t \tanh \dfrac{1}{t}$
Work Step by Step
Since, $\dfrac{d}{dx} (\tanh x)=sech^2 x$
As we are given that $y= t^2 \tanh t^{-1}$
Then, on differentiating , we have
$\dfrac{dy}{dt}= t^2 sech^2 t^{-1} (-t^{-2}) +\tanh t^{-1} (2t)=-sech^2 \dfrac{1}{t}+2t \tanh \dfrac{1}{t}$
