If $f\colon \mathbb{R} \to \mathbb{R}$ is such that $f(x+y)=f(x)f(y)$ and $f'(0)$ exists, then $f$ is differentiable on $\mathbb R$

Question

If $f\colon \mathbb{R} \to \mathbb{R}$ is such that $f(x+y)=f(x)f(y)$ and $f'(0)$ exists, then $f$ is differentiable on $\mathbb R$.

I know how to show it is continuous but no clue how to show differentiable.

Answer 1

Suppose we want to show it is differentiable at a point $p\in\mathbb R$. Note that the given condition implies that $f(0)=1$. (Or $f(0)=0$, but this gives the trivial solution $f=0$).

Then $$\lim_{h\rightarrow 0} \frac{f(p+h)-f(p)}{h} = \lim_{h\rightarrow 0} \frac{f(p)f(h)-f(p)}{h} = f(p)\left(\lim_{h\rightarrow 0}\frac{f(h)-1}{h}\right)=f(p)f'(0).$$