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: 66


$$ y'= 24x+47.$$

Work Step by Step

Taking the $\ln $ on both sides of the equation, we get $$\ln y= \ln (3x+5)(4x+9)$$ Then using the properties of $\ln $, we can write $$\ln y= \ln (3x+5)+\ln(4x+9).$$ Now taking the derivative for the above equation, we have $$\frac{y'}{y}= \frac{3}{3x+5}+ \frac{4}{4x+9},$$ Hence $ y'$ is given by $$ y'=y\left( \frac{3}{3x+5}+ \frac{4}{4x+9}\right)=(3x+5)(4x+9)\left( \frac{3}{3x+5}+ \frac{4}{4x+9}\right)\\ 12x+27+12x+20=24x+47.$$
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