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: 5


$\frac{dy}{dx} =\frac{cos x}{2 \sqrt {sinx}}$

Work Step by Step

$\frac{dy}{du} = \frac{d(\sqrt u)}{du} = \frac{1}{2\sqrt u}$ and, $\frac{du}{dx} = \frac{d(sin x)}{dx} = cos x$ So, $\frac{dy}{dx} = \frac{dy}{du}*\frac{du}{dx}$ $\frac{dy}{dx} =\frac{cos x}{2 \sqrt {u}}$ $\frac{dy}{dx} =\frac{cos x}{2 \sqrt {sinx}}$
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