Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.5 The Chain Rule - 2.5 Exercises: 1

Answer

$\dfrac{dy}{dx}=\dfrac{1}{3}(1+4x)^{-2/3}$

Work Step by Step

Identify the inner function and the outer function $y = \sqrt[3] {1+4x}=(1+4x)^{1/3}$ Outer function: $y=u^{1/3}$ Inner function: $u=1+4x$ $y'=\dfrac{1}{3}u^{-2/3}$ $u'=4$ $\dfrac{dy}{dx}=y'u'=\dfrac{1}{3}(1+4x)^{-2/3}(4)$ $\dfrac{dy}{dx}=\dfrac{1}{3}(1+4x)^{-2/3}$
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