## Calculus: Early Transcendentals 8th Edition

$f'(x) = \frac{2x-2}{x^2-2x}$ The domain is $~~(-\infty, 0) \cup (2, \infty)$
$f(x) = ln(x^2-2x)$ $f'(x) = \frac{1}{x^2-2x}\cdot \frac{d}{dx}(x^2-2x)$ $f'(x) = \frac{2x-2}{x^2-2x}$ 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 $~~(-\infty, 0) \cup (2, \infty)$