Calculus 8th Edition

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

Chapter 2 - Derivatives - 2.3 Differentiation Formulas - 2.3 Exercises: 14

Answer

$y'=\frac{4}{3}\sqrt[3] x+\frac{2}{3}x^{-2/3}$

Work Step by Step

$y=\sqrt[3] x(2+x)$ Use the product rule. $y'=\sqrt[3] x(1)+(2+x)(\frac{1}{3}x^{-2/3})$ Clean up and simplify. $y'=\sqrt[3] x+\frac{2}{3}x^{-2/3}+\frac{1}{3}\sqrt[3] x$ $y'=\frac{4}{3}\sqrt[3] x+\frac{2}{3}x^{-2/3}$
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