Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 386: 33


$$ f'(x) =-\frac{1}{|x|\sqrt{x^2-1}\csc^{-1}x} .$$

Work Step by Step

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