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: 67


$$\frac{dy}{dt}=6\tan(\sin^3t)\sec^2(\sin^3t)\sin^2t\cos t$$

Work Step by Step

$$\frac{dy}{dt}=\frac{d}{dt}\tan^2(\sin^3t)$$ Following the Chain Rule: $$\frac{dy}{dt}=2\tan(\sin^3t)\frac{d}{dt}(\tan(\sin^3t))$$ $$\frac{dy}{dt}=2\tan(\sin^3t)\sec^2(\sin^3t)\frac{d}{dt}(\sin^3t)$$ $$\frac{dy}{dt}=2\tan(\sin^3t)\sec^2(\sin^3t)3\sin^2t\frac{d}{dt}(\sin t)$$ $$\frac{dy}{dt}=6\tan(\sin^3t)\sec^2(\sin^3t)\sin^2t\cos t$$
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