Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.6 - The Chain Rule - Exercises 3.6 - Page 149: 64



Work Step by Step

y=$\frac{1}{6}[1+cos^2(7t)]^3$ Apply the chain rule: $\frac{dy}{dt}=\frac{3}{6}[1+cos^2(7t)]^2.2cos(7t)(-sin(7t)(7))$ =-7$[1+cos^2(7t)]^2(cos(7t)sin(7t))$
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