Answer
$f'(x) = \frac{2x-2}{x^2-2x}$
The domain is $~~(-\infty, 0) \cup (2, \infty)$
Work Step by Step
$f(x) = ln(x^2-2x)$
$f'(x) = \frac{1}{x^2-2x}\cdot \frac{d}{dx}(x^2-2x)$
$f'(x) = \frac{2x-2}{x^2-2x}$
When we consider the function $f(x) = ln(x^2-2x)$, it is required that $(x^2-2x) \gt 0$
Then:
$x(x-2) \gt 0$
$x \lt 0~~$ or $~~x \gt 2$
The domain is $~~(-\infty, 0) \cup (2, \infty)$
