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: 29

Answer

$\frac{5}{4}.$

Work Step by Step

Since $ y=\sin^{-1}x $, then $$ y'= \frac{1}{\sqrt{1-x^2}}.$$ Now, we have $$ y'(3/5)= \frac{1}{\sqrt{1-(3/5)^2}}=\frac{1}{(4/5)}=\frac{5}{4}.$$
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