# How do you differentiate  f(x)=ln(x^2-x) using the chain rule.?

Mar 4, 2016

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

#### Explanation:

The chain rule for the specific case of a natural logarithm function can be derived as follows:

$\frac{d}{\mathrm{dx}} \ln \left(x\right) = \frac{1}{x} \text{ "=>" } \frac{d}{\mathrm{dx}} \ln \left(g \left(x\right)\right) = \frac{1}{g \left(x\right)} \cdot g ' \left(x\right)$

So, for $f \left(x\right) = \ln \left({x}^{2} - x\right)$, we see that

$f ' \left(x\right) = \frac{1}{{x}^{2} - x} \left(\frac{d}{\mathrm{dx}} \left({x}^{2} - x\right)\right)$

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

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