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: 20

Answer

$F'(t)=-6(3t-1)^{4}(2t+1)^{-4} / +12(2t+1)^{-3}(3t-1)^{3}$

Work Step by Step

$F(t)=(3t-1)^{4}(2t+1)^{-3}$ Differentiate using the product rule: $F'(t)=[(3t-1)^{4}][(2t+1)^{-3}]'+[(2t+1)^{-3}][(3t-1)^{4}]'=...$ Use the chain rule to find $[(2t+1)^{-3}]'$ and $[(3t-1)^{4}]'$: $...=[(3t-1)^{4}][-3(2t+1)^{-4}(2t+1)']+[(2t+1)^{-3}][4(3t-1)^{3}(3t-1)']=...$ $...=[(3t-1)^{4}][-3(2t+1)^{-4}(2)]+[(2t+1)^{-3}][4(3t-1)^{3}(3)]=...$ $...=-6(3t-1)^{4}(2t+1)^{-4} / +12(2t+1)^{-3}(3t-1)^{3}$
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