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: 7

Answer

$\frac{dy}{dx} = (2x \pi)sec^{2}(\pi x^{2})$

Work Step by Step

$\frac{dy}{du} = \frac{d (tan u)}{du} = sec^{2}u$ and, $\frac{du}{dx} = \frac{d (\pi x^{2})}{dx} = 2x \pi$ So, $\frac{dy}{dx} = \frac{dy}{du} * \frac{du}{dx}$ $\frac{dy}{dx} = (2x \pi) sec^{2} u$ $\frac{dy}{dx} = (2x \pi)sec^{2}(\pi x^{2})$
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