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: 52

Answer

$$\frac{dy}{dt}==2\pi\sec^2\pi t\tan\pi t$$

Work Step by Step

According to the Chain Rule: $$\frac{dy}{dt}=\frac{d}{dt}(\sec^2\pi t)$$ $$\frac{dy}{dt}=2\sec\pi t\frac{d}{dt}(\sec \pi t)$$ $$\frac{dy}{dt}=2\sec\pi t(\sec\pi t\tan\pi t)\frac{d}{dt}(\pi t)$$ $$\frac{dy}{dt}=2\sec^2\pi t\tan\pi t\times(\pi)=2\pi\sec^2\pi t\tan\pi t$$
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