Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.7 The Chain Rule - Exercises - Page 146: 55


$$ y'=3\tan^2x \sec^2 x+3x^2\sec^2 x^3.$$

Work Step by Step

Recall that $(\tan x)'=\sec^2 x$. Recall that $(x^n)'=nx^{n-1}$ Since $ y=\tan^3 x+\tan x^3$, by using the chain rule, the derivative $ y'$ is given by $$ y'=3\tan^2x \sec^2 x+3x^2\sec^2 x^3.$$
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