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: 35


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

Work Step by Step

Since $ y=cos^{-1}(x^2) $, then $$ y'=- \frac{1}{ \sqrt{1-(x^2)^2}}(x^2)'=- \frac{2x}{ \sqrt{1-(x^2)^2}}.$$
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