Answer
$\frac{f'(x)}{f(x)} = \frac{1}{(x-a)}+\frac{1}{(x-b)}+\frac{1}{(x-c)}$
Work Step by Step
$f(x) = (x-a)(x-b)(x-c)$
Let $g(x) = (x-a)(x-b)$ and let $h(x) = (x-c)$
Then $f'(x) = g'(x)~h(x)+g(x)~h'(x)$
We can find $f'(x)$:
$f'(x) = [(x-b)+(x-a)]\cdot (x-c)+(x-a)(x-b)$
$f'(x) = (x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)$
Then:
$\frac{f'(x)}{f(x)} = \frac{(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)}{(x-a)(x-b)(x-c)}$
$\frac{f'(x)}{f(x)} = \frac{(x-b)(x-c)}{(x-a)(x-b)(x-c)}+\frac{(x-a)(x-c)}{(x-a)(x-b)(x-c)}+\frac{(x-a)(x-b)}{(x-a)(x-b)(x-c)}$
$\frac{f'(x)}{f(x)} = \frac{1}{(x-a)}+\frac{1}{(x-b)}+\frac{1}{(x-c)}$
