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: 51


$$\frac{dy}{dt}=\pi\sin(2\pi t-4)$$

Work Step by Step

According to the Chain Rule: $$\frac{dy}{dt}=\frac{d}{dt}\sin^2(\pi t-2)=2\sin(\pi t-2)\frac{d}{dt}\sin(\pi t-2)$$ $$\frac{dy}{dt}=2\sin(\pi t-2)\cos(\pi t-2)\frac{d}{dt}(\pi t-2)$$ $$\frac{dy}{dt}=2\sin(\pi t-2)\cos(\pi t-2)(\pi\times1-0)$$ $$\frac{dy}{dt}=2\pi\sin(\pi t-2)\cos(\pi t-2)$$ Recall the identity $2\sin A\cos A=\sin2A$: $$\frac{dy}{dt}=\pi\sin(2\pi t-4)$$
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