Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.6 - Derivatives of Logarithmic Functions - 3.6 Exercises - Page 223: 13



Work Step by Step

$G(y)=\ln \frac{(2y+1)^5}{\sqrt{y^2+1}}\\ G(y)=5\ln(2y+1)-\frac{1}{2}\ln(y^2+1)\\ G'(y)=\frac{5\times2}{2y+1}-\frac{1\times2y}{2\times(y^2+1)}\\ G'(y)=\frac{10}{2y+1}-\frac{y}{(y^2+1)}\\ G'(y)=\frac{10y^2+10-2y^2-y}{(2y+1)(y^2+1)}\\ G'(y)=\frac{8y^2+10-y}{(2y+1)(y^2+1)}$
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