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: 15


$sech^2 \sqrt t + \dfrac{\tanh \sqrt t}{\sqrt t}$

Work Step by Step

Since, $\dfrac{d}{dx} (\tanh x)=sech^2 x$ As we are given that $y=2 t^{1/2} \tanh (t^{1/2})$ Then, on differentiating , we have $\dfrac{dy}{dt}= 2[t^{1/2} sech^2 (t^{1/2})( \dfrac{1}{2}t^{-1/2} )+ \tanh (t^{1/2}) (\dfrac{1}{2}t^{-1/2})] = sech^2 \sqrt t+ t^{-1/2} \tanh \sqrt t=sech^2 \sqrt t + \dfrac{\tanh \sqrt t}{\sqrt t}$
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.