Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.7 - Hyperbolic Functions - Exercises 7.7 - Page 430: 16


$-sech^2 \dfrac{1}{t}+2t \tanh \dfrac{1}{t}$

Work Step by Step

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, $=-sech^2 \dfrac{1}{t}+2t \tanh \dfrac{1}{t}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.