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: 33


$$ y'= \frac{7}{\sqrt{1-(7x)^2}}.$$

Work Step by Step

Since $ y=\sin^{-1}7x $, then $$ y'= \frac{1}{\sqrt{1-(7x)^2}}(7x)'=\frac{7}{\sqrt{1-(7x)^2}}.$$
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