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: 19

Answer

$$h'(t)=\frac{2(2t^2-1)^2(20t^2+18t-1)}{3(t+1)^{1/3}}$$

Work Step by Step

$$h'(t)=((t+1)^{2/3}(2t^2-1)^3)'= ((t+1)^{2/3})'(2t^2-1)^3+(t+1)^{2/3}((2t^2-1)^3)'=\frac{2}{3}(t+1)^{2/3-1}(t+1)'(2t^2-1)^3+(t+1)^{2/3}\cdot3(2t^2-1)^2(2t^2-1)'= \frac{2}{3}\frac{1}{(t+1)^{1/3}}\cdot1(2t^2-1)^3+3(t+1)^{2/3}(2t^2-1)^2\cdot4t= \frac{2(2t^2-1)^3+3\cdot12t(t+1)(2t^2-1)^2}{3(t+1)^{1/3}}= \frac{2(2t^2-1)^2(2t^2-1+18t(t+1))}{3(t+1)^3}= \frac{2(2t^2-1)^2(20t^2+18t-1)}{3(t+1)^{1/3}}$$
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