Chapter 7: Transcendental Functions - Section 7.7 - Hyperbolic Functions - Exercises 7.7 - Page 430: 30

$1-2t\coth^{-1}t$

Given: $y=(1-t^2) \coth^{-1} t$ Since, $\dfrac{d (\coth^{-1} x)}{dx}=\dfrac{1}{1-x^2}$ Apply product rule to get the differentiation. Thus, $\dfrac{dy}{dt}=(1-t^2) \dfrac{1}{1-t^2} +\coth^{-1}t (-2t)$ or, $\dfrac{dy}{dt}=1-2t\coth^{-1}t$