Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.4 Exercises: 13

Answer

$y'=\dfrac{4x}{\sqrt[3]{(6x^2+1)^2}}$

Work Step by Step

$u=(1+6x^2)$; $\dfrac{du}{dx}=(12x)$ $y=\sqrt[3]{u};\dfrac{dy}{du}=\dfrac{1}{3\sqrt[3]{u^2}}$ $\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}=(12x)(\dfrac{1}{3\sqrt[3]{(6x^2+1)^2}})=\dfrac{4x}{\sqrt[3]{(6x^2+1)^2}}$
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