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: 68


$$ y''=32 \cos (8x+18).$$

Work Step by Step

Recall that $(\sin x)'=\cos x$. Given $ y=\sin^2 (4x+9)$, then, by using the chain rule, we have $$ y'=2\sin(4x+9) \cos (4x+9) (4)=8\sin(4x+9) \cos (4x+9) .$$ One can simplify $ y'$ using the fact that $2\sin x \cos x=\sin2x $ as follows $$ y'=4\sin (8x+18)$$ and, by using the chain rule, the second derivative $ y''$ is given by $$ y''=4\cos (8x+18) (8)=32 \cos (8x+18).$$
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