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: 24



Work Step by Step

Use the hyperbolic functions formulae such as: $ \sinh x= \dfrac{e^x-e^{-x}}{2}$ Thus, $\text{csch} x=\dfrac{1}{\sinh x}=\dfrac{2}{e^x -e^{-x}}$ We are given $y=(4x^2 -1) \text{csch}(\ln 2x)$ or, $y=(4x^2 -1)\dfrac{2}{e^{\ln 2x} - e^{-\ln 2x}}= (4x^2 -1)\dfrac{2}{e^{\ln 2x}+e^{\ln (2x)^{-1}}}=(4x^2 -1)\dfrac{4x}{4x^2-1}=4x$ Now, $\dfrac{dy}{dx}=4$
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.