Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.9 Hyperbolic Functions - Exercises - Page 384: 22


$$ y'=-\frac{csch^2c}{\coth x} .$$

Work Step by Step

Recall that $(\ln x)'=\dfrac{1}{x}$ Recall that $(\coth x)'=-\csc^2 x$ Since $ y=\ln(\coth x)$, then the derivative, by using the chain rule, is given by $$ y'=\frac{1}{\coth x} (\coth x)'=-\frac{csch^2c}{\coth x} .$$
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.