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: 57


$$\frac{dy}{dt}=-\pi e^{\cos^2(\pi t-1)}\sin(2\pi t-2)$$

Work Step by Step

$$\frac{dy}{dt}=\frac{d}{dt}e^{\cos^2(\pi t-1)}$$ According to the Chain Rule: $$\frac{dy}{dt}=e^{\cos^2(\pi t-1)}\frac{d}{dt}\Big(\cos^2(\pi t-1)\Big)'$$ $$\frac{dy}{dt}=e^{\cos^2(\pi t-1)}2\cos(\pi t-1)\frac{d}{dt}(\cos(\pi t-1))$$ $$\frac{dy}{dt}=e^{\cos^2(\pi t-1)}2\cos(\pi t-1)(-\sin(\pi t-1))\frac{d}{dt}(\pi t-1)$$ $$\frac{dy}{dt}=-e^{\cos^2(\pi t-1)}2\cos(\pi t-1)\sin(\pi t-1)(\pi-0)$$ $$\frac{dy}{dt}=-\pi e^{\cos^2(\pi t-1)}2\cos(\pi t-1)\sin(\pi t-1)$$ Recall the identity $2\cos\alpha\sin\alpha=\sin2\alpha$: $$\frac{dy}{dt}=-\pi e^{\cos^2(\pi t-1)}\sin(2\pi t-2)$$
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