Answer
$g'(x) = x~f'(x)+f(x)$
Work Step by Step
$g(x) = x~f(x)$
We can find an expression for $g'(x)$:
$g'(x) = \lim\limits_{h \to 0}\frac{g(x+h)-g(x)}{h}$
$g'(x) = \lim\limits_{h \to 0}\frac{(x+h)~f(x+h)-x~f(x)}{h}$
$g'(x) = \lim\limits_{h \to 0}\frac{x~f(x+h)-x~f(x)+h~f(x+h)}{h}$
$g'(x) = \lim\limits_{h \to 0}\frac{x~f(x+h)-x~f(x)}{h}+\lim\limits_{h \to 0}\frac{h~f(x+h)}{h}$
$g'(x) = x~\Big(\lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}\Big)+\lim\limits_{h \to 0}f(x+h)$
$g'(x) = x~f'(x)+f(x)$
