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: 17

Answer

$\frac{dy}{dx} = 3tan^{2} x \times sec^{2} x$

Work Step by Step

$y = u^{3}$ where, $u = tan x$ Now, $\frac{dy}{du} = 3u^{2}$ and, $\frac{du}{dx} =sec^{2} x$ So, $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$ $\frac{dy}{dx} = 3u^{2} \times sec^{2} x$ $\frac{dy}{dx} = 3tan^{2} x \times sec^{2} x$
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