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 - Page 136: 10



Work Step by Step

$u=9t+2$; $\dfrac{du}{dt}=9$ $f(u)=u^{\frac{2}{3}};\dfrac{d}{du}f(u)=\dfrac{2}{3\sqrt[3]{u}}$ $\dfrac{d}{dt}f(t)=\dfrac{d}{du}f(u)\times\dfrac{du}{dt}=\dfrac{6}{\sqrt[3]{u}}=\dfrac{6}{\sqrt[3]{9t+2}}$
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