Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.4 Derivatives of Logarithmic Functions - 6.4 Exercises: 13

Answer

$\frac{10}{2y+1}-\frac{y}{y^2+1}$

Work Step by Step

$\frac{d}{dy}\ln\frac{(2y+1)^5}{\sqrt{y^2+1}}$ In this case, it is much easier to simplify the expression using the logarithm laws (Theorem 3 on page 422) before taking the derivative. $=\frac{d}{dy}(\ln(2y+1)^5-\ln\sqrt{y^2+1})$ $=\frac{d}{dy}(5\ln(2y+1)-\frac{1}{2}\ln(y^2+1))$ $=5*\frac{1}{2y+1}\frac{d}{dy}(2y+1)-\frac{1}{2}*\frac{1}{y^2+1}*\frac{d}{dy}(y^2+1)$ $=5*\frac{1}{2y+1}*2-\frac{1}{2}*\frac{1}{y^2+1}*2y$ $=\frac{10}{2y+1}-\frac{y}{y^2+1}$
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