Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 68


$$ y' =\frac{x(x+1)^3}{(3x-1)^2}\left( \frac{1}{ x}- \frac{3}{ x+1}-\frac{6}{3 x-1}\right).$$

Work Step by Step

Taking the $\ln $ of both sides of the equation, we get $$\ln y= \ln \frac{x(x+1)^3}{(3x-1)^2}$$ Then using the properties of $\ln $, we have $$\ln y= \ln x+3\ln (x+1)-2\ln(3x-1).$$ Now taking the derivative for the above equation, we have $$\frac{y'}{y}= \frac{1}{ x}+ \frac{3}{ x+1}-\frac{6}{3 x-1},$$ Hence $ y'$ is given by $$ y'=y\left( \frac{1}{ x}+ \frac{3}{ x+1}-\frac{6}{3 x-1}\right)\\ =\frac{x(x+1)^3}{(3x-1)^2}\left( \frac{1}{ x}+ \frac{3}{ x+1}-\frac{6}{3 x-1}\right).$$
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