Single Variable Calculus, Chapter 7, 7.2-2, Section 7.2-2, Problem 38

Differentiate $f(x) = \ln (x^2 - 2x)$ and find the domain of $f$.



$\begin{equation} \begin{aligned} \text{if } f(x) =& \ln (x^2 - 2x), \text{ then} \\ \\ f'(x) =& \frac{\displaystyle \frac{d}{dx} (x^2 - 2x) }{x^2 -2x} \\ \\ f'(x) =& \frac{2x - 2}{x^2 - 2x} \end{aligned} \end{equation}$


We know that $f(x)$ is a natural logarithmic function that is defined for $x^2 - 2x > 0$


$\begin{equation} \begin{aligned} & x( x - 2) > 0 \to (x > 0) \bigcap (x - 2 > 0) \\ \\ & \text{Since } (x > 0) \bigcap (x - 2 > 0) \\ \\ & \text{We have,} \\ \\ & x < 0 \text{ and } x > 2 \end{aligned} \end{equation}$


Therefore, the domain of $f$ is

$(- \infty, 0) \bigcup (2, \infty)$