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 163: 45


$$2\cos2x \cos^2x-\sin^2 2x$$

Work Step by Step

Recall the product rule: $(uv)'=u'v+uv'$ Recall that $(\sin x)'=\cos x$. Recall that $(\cos x)'=-\sin x$. Since $ y=\sin 2x \cos^2x $, then by the product rule, the derivative $ y'$ is given by $$ y'=\cos 2x (2) \cos^2x+\sin 2x (2)\cos x (-\sin x)\\=2\cos2x \cos^2x-2\sin 2x \cos x \sin x.$$ Using the formula $\sin 2x=2\cos x \sin x$, we write: $$2\cos2x \cos^2x-\sin^2 2x$$
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