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: 29

Answer

$$H'(r)=\frac{2(r^2-1)^2(r^2+3r+5)}{(2r+1)^6}$$

Work Step by Step

$$H'(r)=\left(\frac{(r^2-1)^3}{(2r+1)^5}\right)'=\frac{((r^2-1)^3)'(2r+1)^5-(r^2-1)^3((2r+1)^5)'}{(2r+1)^{10}}=\frac{3(r^2-1)^2(r^2-1)'(2r+1)^5-(r^2-1)^3\cdot5(2r+1)^4(2r+1)'}{(2r+1)^{10}}= \frac{3(r^2-1)^2\cdot2r(2r+1)^5-5(r^2-1)^3(2r+1)^4\cdot2}{(2r+1)^{10}}= \frac{6r(r^2-1)^2(2r+1)^5-10(r^2-1)^3(2r+1)^4}{(2r+1)^{10}}= \frac{2(r^2-1)^2(2r+1)^4(3r(2r+1)-5(r^2-1))}{(2r+1)^{10}}= \frac{2(r^2-1)^2(6r^2+3r-5r^2+5)}{(2r+1)^6}= \frac{2(r^2-1)^2(r^2+3r+5)}{(2r+1)^6}$$
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