# How do you find the derivative of y= ln ((x^2 (x+1))/ (x+2)^3)?

Jan 9, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5 x + 4}{x \left(x + 1\right) \left(x + 2\right)}$

#### Explanation:

$y = \ln \left\{\frac{{x}^{2} \left(x + 1\right)}{x + 2} ^ 3\right\}$

Using the Rules of $L o g$ function, we have,

$y = \ln {x}^{2} + \ln \left(x + 1\right) - \ln {\left(x + 2\right)}^{3}$

$= 2 \ln x + \ln \left(x + 1\right) - 3 \ln \left(x + 2\right)$

Diff.ing the L.H.S., we will use the Chain Rule.

E.g., $\frac{d}{\mathrm{dx}} \left\{\ln \left(x + 1\right)\right\} = \frac{1}{x + 1} \frac{d}{\mathrm{dx}} \left(x + 1\right) = \frac{1}{x + 1}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left(\frac{1}{x}\right) + \frac{1}{x + 1} - \frac{3}{x + 2}$

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

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5 x + 4}{x \left(x + 1\right) \left(x + 2\right)}$