# How do I find the derivative of ln(x-2)/(x-2)?

Jan 30, 2016

$\frac{1 - \ln \left(x - 2\right)}{x - 2} ^ 2$

#### Explanation:

Use the quotient rule, which states that

$\frac{d}{\mathrm{dx}} \left[\frac{f \left(x\right)}{g \left(x\right)}\right] = \frac{f ' \left(x\right) g \left(x\right) - g ' \left(x\right) f \left(x\right)}{g \left(x\right)} ^ 2$

Applying this to $\ln \frac{x - 2}{x - 2}$, we see that\

$\frac{d}{\mathrm{dx}} \left[\ln \frac{x - 2}{x - 2}\right] = \frac{\left(x - 2\right) \frac{d}{\mathrm{dx}} \left[\ln \left(x - 2\right)\right] - \ln \left(x - 2\right) \frac{d}{\mathrm{dx}} \left[x - 2\right]}{x - 2} ^ 2$

The respective derivatives are:

$\frac{d}{\mathrm{dx}} \left[\ln \left(x - 2\right)\right] = \frac{1}{x - 2} \frac{d}{\mathrm{dx}} \left[x - 2\right] = \frac{1}{x - 2}$

$\frac{d}{\mathrm{dx}} \left[x - 2\right] = 1$

Yielding:

$\frac{d}{\mathrm{dx}} \left[\ln \frac{x - 2}{x - 2}\right] = \frac{\left(x - 2\right) \frac{1}{x - 2} - \ln \left(x - 2\right)}{x - 2} ^ 2$

$= \frac{1 - \ln \left(x - 2\right)}{x - 2} ^ 2$