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: 71


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

Work Step by Step

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