Prove $ \frac{d}{dt}(f *g)(t) = \int_0^tg(t-s)f^{'}(s)ds \; \; + f(0)g(t) $

Question

Given that $f,g$ are continuous real and differentiable functions on $[0,T]$ and

$$ (f*g)(t) := \int_0^tf(t-s)g(s)ds $$

Prove that:

$$ \frac{d}{dt}(f *g)(t) = \int_0^tg(t-s)f^{'}(s)ds \; \; + f(0)g(t) $$