# How do you differentiate y=ln[(x+9)^6(x+6)^2(x+5)^3]?

Sep 18, 2017

First use properties of logarithms to expand this as a sum and then differentiate to get $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6}{x + 9} + \frac{2}{x + 6} + \frac{3}{x + 5}$. If you add these fractions, you can also say $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{11 {x}^{2} + 139 x + 432}{{x}^{3} + 20 {x}^{2} + 129 x + 270}$.

#### Explanation:

Properties of logarithms allow us to write $y = \ln \left[{\left(x + 9\right)}^{6} {\left(x + 6\right)}^{2} {\left(x + 5\right)}^{3}\right] = \ln \left[{\left(x + 9\right)}^{6}\right] + \ln \left[{\left(x + 6\right)}^{2}\right] + \ln \left[{\left(x + 5\right)}^{3}\right] .$

Continuing, this is also equal to

$y = 6 \ln \left(x + 9\right) + 2 \ln \left(x + 6\right) + 3 \ln \left(x + 5\right)$.

This is now easy to differentiate (and the use of the Chain Rule is trivial) to get

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6}{x + 9} + \frac{2}{x + 6} + \frac{3}{x + 5}$.

Getting a common denominator gives

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 \left(x + 6\right) \left(x + 5\right) + 2 \left(x + 9\right) \left(x + 5\right) + 3 \left(x + 9\right) \left(x + 6\right)}{\left(x + 9\right) \left(x + 6\right) \left(x + 5\right)}$.

We continue to simplify:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 {x}^{2} + 66 x + 180 + 2 {x}^{2} + 28 x + 90 + 3 {x}^{2} + 45 x + 162}{\left({x}^{2} + 15 x + 54\right) \left(x + 5\right)}$

$= \frac{11 {x}^{2} + 139 x + 432}{{x}^{3} + 15 {x}^{2} + 54 x + 5 {x}^{2} + 75 x + 270}$

So

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{11 {x}^{2} + 139 x + 432}{{x}^{3} + 20 {x}^{2} + 129 x + 270}$