University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.3 - Hyperbolic Functions - Exercises - Page 417: 30



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: $\dfrac{dy}{dt}=(1-t^2) \dfrac{1}{1-t^2} +\coth^{-1}t (-2t)=1-2t\coth^{-1}t$
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