suppose f is a continues function on point x_0

prove that g(x)=(x-x_0)*f(x) differentiable on x_0??

calculate g'(x_0)

i tried to think like this:

if f(x) is continues on x_0 then lim f(x) as x->x_0 equals f(x_0)

mvt says f'(c)=[f(a)-f(b)]

cauchys mvt says f'(c)/g'(c)=[f(a)-f(b)]/[g(a)-g(b)]

??