Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises - Page 204: 19



Work Step by Step

$h(t)=(t+1)^{2/3}(2t^{2}-1)^{3}$ Differentiate using the product rule: $h'(t)=[(t+1)^{2/3}][(2t^{2}-1)^{3}]'+[(2t^{2}-1)^{3}][(t+1)^{2/3}]'=...$ Use the chain rule to find $[(2t^{2}-1)^{3}]'$ and $[(t+1)^{2/3}]'$: $...=[(t+1)^{2/3}][3(2t^{2}-1)^{2}(2t^{2}-1)']+[(2t^{2}-1)^{3}][\dfrac{2}{3}(t+1)^{-1/3}(t+1)']=...$ $...=[(t+1)^{2/3}][3(2t^{2}-1)^{2}(4t)]+[(2t^{2}-1)^{3}][\dfrac{2}{3}(t+1)^{-1/3}(1)]=...$ $...=12t(t+1)^{2/3}(2t^{2}-1)^{2}+\dfrac{2}{3}(t+1)^{-1/3}(2t^{2}-1)^{3}$
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