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: 20

Answer

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

Work Step by Step

$$F'(t)=((3t-1)^4(2t+1)^{-3})'=((3t-1)^4)'\cdot(2t+1)^{-3})+(3t-1)^4((2t+1)^{-3})'=4(3t-1)^{4-1}\cdot(3t-1)'\cdot(2t+1)^{-3}+(3t-1)^4\cdot(-3)(2t+1)^{-3-1}\cdot)(2t+1)'= 4(3t-1)^3\cdot3(2t+1)^{-3}+-3(3t-1)^4(2t+1)^{-4}\cdot2= \frac{12(3t-1)^3}{(2t+1)^3}-\frac{6(3t+1)^4}{(2t+1)^4}= \frac{12(2t+1)(3t-1)^3-6(3t+1)^4}{(2t+1)^4}= \frac{6(3t-1)^3(2(2t+1)-(3t+1))}{(2t+1)^4}= \frac{6(3t-1)^3(t+3)}{(2t+1)^4}$$
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