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 - Page 158: 15



Work Step by Step

$$h'(v)=(v \sqrt[3]{1+v^2})'=v'\cdot\sqrt[3]{1+v^2}+v\cdot(\sqrt[3]{1+v^2})'= 1\cdot\sqrt[3]{1+v^2}+v\cdot\frac{1}{3}(1+v^2)^{\frac{1}{3}-1}(1+v^2)'= \sqrt[3]{1+v^2}+\frac{1}{3}v\frac{1}{\sqrt[3]{1+v^2}^2}\cdot2v= \frac{3(1+v^2)+2v^2}{3\sqrt[3]{1+v^2}^2}=\frac{3+5v^2}{3\sqrt[3]{1+v^2}^2}$$
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