Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 374: 40


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

Work Step by Step

Since $ y=\csc^{-1}(x^{-1})$, then the derivative $ y'$ is given by $$ y'= -\frac{1}{|x^{-1}|\sqrt{(x^{-1})^2-1}} (x^{-1})'= \frac{x^{-2}}{|x^{-1}|\sqrt{(x^{-1})^2-1}} .$$ This can be simplified to: $$y'=\dfrac{1}{\sqrt{1-x^2}}$$
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.