Answer
$\frac{5}{4}.$
Work Step by Step
Since $ y=\sin^{-1}x $, then
$$ y'= \frac{1}{\sqrt{1-x^2}}.$$
Now, we have
$$ y'(3/5)= \frac{1}{\sqrt{1-(3/5)^2}}=\frac{1}{(4/5)}=\frac{5}{4}.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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