## Thomas' Calculus 13th Edition

$-sech^2 \dfrac{1}{t}+2t \tanh \dfrac{1}{t}$
Given: $y= t^2 \tanh t^{-1}$ Since, $\dfrac{d}{dx} (\tanh x)=sech^2 x$ This implies $\dfrac{dy}{dt}= (t^2) sech^2 (t^{-1}) (-t^{-2}) +(2t)\tanh (t^{-1} )$ or, $=-sech^2 \dfrac{1}{t}+2t \tanh \dfrac{1}{t}$