Calculus (3rd Edition)

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

Chapter 3 - Differentiation - Chapter Review Exercises - Page 164: 67


$$ y''= 2\sec x^2+4x^2 \sec x^2 \tan x^2.$$

Work Step by Step

Recall that $(\tan x)'=\sec^2 x$. Recall that $(x^n)'=nx^{n-1}$ Given $ y=\tan x^2$, then, by using the chain rule, we have $$ y'=(2x)\sec x^2=2x\sec x^2,$$ and, by using the product rule, the second derivative $ y''$ is given by $$ y''=2\sec x^2+2x (2x) \sec x^2 \tan x^2 \\=2\sec x^2+4x^2 \sec x^2 \tan x^2.$$
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