Answer
$-\frac{20}{3}.$
Work Step by Step
Since $ y=\cos^{-1}4x $, then
$$ y'= -\frac{1}{\sqrt{1-(4x)^2}}(4x)'=-\frac{4}{\sqrt{1-(4x)^2}}.$$
Now, we have
$$ y'(1/5)=-\frac{4}{\sqrt{1-(4/5)^2}}=-\frac{4}{\sqrt{ 9/25}} =-\frac{20}{3}.$$
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: 32
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