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: 6

Answer

$\frac{dy}{dx} = (1+sin x)cos (x-cos x)$

Work Step by Step

$\frac{dy}{du} = \frac{d(sin u)}{du} = cos u$ and, $\frac{du}{dx} = \frac{d (x-cos x)}{dx} = 1+sin x$ So, $\frac{dy}{dx} = \frac{dy}{du}*\frac{du}{dx}$ $\frac{dy}{dx} = cos u (1+sin x)$ $\frac{dy}{dx} = (1+sin x)cos (x-cos x)$
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