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: 61



Work Step by Step

$$\frac{dy}{dt}=\frac{d}{dt}(\sin(\cos(2t-5)))$$ Following the Chain Rule: $$\frac{dy}{dt}=\cos(\cos(2t-5))\frac{d}{dt}(\cos(2t-5))$$ $$\frac{dy}{dt}=\cos(\cos(2t-5))(-\sin(2t-5))\frac{d}{dt}(2t-5)$$ $$\frac{dy}{dt}=-\cos(\cos(2t-5))\sin(2t-5)(2\times1-0)$$ $$\frac{dy}{dt}=-2\cos(\cos(2t-5))\sin(2t-5)$$
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