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: 2

Answer

$\frac{dy}{dx} = 48(64x^{2} +1 - 16x)=48(8x-1)^{2}$

Work Step by Step

$\frac{dy}{du} = \frac{d(2u^{3})}{du} = 6u^{2}$ And, $\frac{du}{dx} = \frac{d(8x-1)}{dx} = 8$ So, $\frac{dy}{dx} = \frac{dy}{du}*\frac{du}{dx}$ $\frac{dy}{dx} = 48u^{2}$ $\frac{dy}{dx} = 48(8x-1)^{2}$ $\frac{dy}{dx} = 48(64x^{2} +1 - 16x)$
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