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: 53



Work Step by Step

$$\frac{dy}{dt}=\frac{d}{dt}(1+\cos2t)^{-4}$$ According to the Chain Rule: $$\frac{dy}{dt}=-4(1+\cos2t)^{-5}\frac{d}{dt}(1+\cos2t)$$ $$\frac{dy}{dt}=-4(1+\cos2t)^{-5}\Big(0-\sin2t\frac{d}{dt}(2t)\Big)$$ $$\frac{dy}{dt}=-4(1+\cos2t)^{-5}(-2\sin2t)$$ $$\frac{dy}{dt}=8\sin2t(1+\cos2t)^{-5}$$
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