Answer
$$ y'= \frac{7}{\sqrt{1-(7x)^2}}.$$
Work Step by Step
Since $ y=\sin^{-1}7x $, then
$$ y'= \frac{1}{\sqrt{1-(7x)^2}}(7x)'=\frac{7}{\sqrt{1-(7x)^2}}.$$
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: 33
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