# Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 65

$$y'= 2x+14.$$

#### Work Step by Step

Taking the $\ln$ on both sides of the equation, we get $$\ln y= \ln (x+5)(x+9)$$ Then using the properties of $\ln$, we can write $$\ln y= \ln (x+5)+\ln(x+9).$$ Now taking the derivative for the above equation, we have $$\frac{y'}{y}= \frac{1}{x+5}+ \frac{1}{x+9},$$ Hence $y'$ is given by $$y'=y\left( \frac{1}{x+5}+ \frac{1}{x+9}\right)=(x+5)(x+9)\left( \frac{1}{x+5}+ \frac{1}{x+9}\right)\\ x+9+x+5=2x+14.$$

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