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: 32

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 $$ y'(1/5)=-\frac{4}{\sqrt{1-(4/5)^2}}=-\frac{4}{\sqrt{ 9/25}} =-\frac{20}{3}.$$
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