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: 70


$$ y'= (2x+1)(4x^2)\sqrt{x-9})\left( \frac{2}{ 2x+1}+\frac{4}{ 2x}+\frac{1}{2}\frac{1}{x-9}\right).$$

Work Step by Step

Taking the $\ln $ of both sides of the equation, we get $$\ln y= \ln((2x+1)(4x^2)\sqrt{x-9})$$ Then using the properties of $\ln $, we have $$\ln y= \ln (2x+1)+\ln (2x)^2+\frac{1}{2}\ln(x-9)\\ =\ln (2x+1)+2\ln (2x)+\frac{1}{2}\ln(x-9).$$ Now taking the derivative for the above equation, we have $$\frac{y'}{y}= \frac{2}{ 2x+1}+\frac{4}{ 2x}+\frac{1}{2}\frac{1}{x-9},$$ Hence $ y'$ is given by $$ y'=y\left( \frac{2}{ 2x+1}+\frac{4}{ 2x}+\frac{1}{2}\frac{1}{x-9}\right)\\ =(2x+1)(4x^2)\sqrt{x-9})\left( \frac{2}{ 2x+1}+\frac{4}{ 2x}+\frac{1}{2}\frac{1}{x-9}\right).$$
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