Answer
$$ y'= \frac{3(\tan^{-1}x)^2}{1+x^2}.$$
Work Step by Step
Since $ y=(\tan^{-1}x)^3$, then the derivative is given by
$$ y'=3(\tan^{-1}x)^2(\tan^{-1}x)'=\frac{3(\tan^{-1}x)^2}{1+x^2}.$$
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: 43
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