University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 158: 64



Work Step by Step

$$\frac{dy}{dt}=\frac{d}{dt}\frac{1}{6}\Big(1+\cos^2(7t)\Big)^3$$ Following the Chain Rule: $$\frac{dy}{dt}=\frac{1}{6}\times3\Big(1+\cos^2(7t)\Big)^2\frac{d}{dt}(1+\cos^2(7t))$$ $$\frac{dy}{dt}=\frac{1}{2}(1+\cos^2(7t))^2(0+2\cos7t\frac{d}{dt}(\cos7t))$$ $$\frac{dy}{dt}=\frac{1}{2}(1+\cos^2(7t))^2(2\cos7t(-\sin7t)\frac{d}{dt}(7t))$$ $$\frac{dy}{dt}=\frac{1}{2}(1+\cos^2(7t))^2(-14\cos7t\sin7t)$$ $$\frac{dy}{dt}=-7\sin7t\cos7t(1+\cos^2(7t))^2$$
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