If f(a+b) = f(a)f(b) then f is exponential

Question

I need to prove that if $f(a+b)=f(a)f(b)$ for any two real numbers $a$, $b$, and $f(0)=f'(0)=1$, then $f(x)=f'(x)$ for all real number $x$.

So here's what I tried but it didn't quite work and I can't find a way to get the desired result;

Let $x,c$ be any real number,

$f(x)=f((\frac{x}{2}+c)+(\frac{x}{2}-c))=f(\frac{x}{2}+c)f(\frac{x}{2}-c)$

$f'(x)=\frac{1}{2}f(\frac{x}{2}+c)f'(\frac{x}{2}-c) + \frac{1}{2}f(\frac{x}{2}-c)f'(\frac{x}{2}+c)$

Substitute $f'(0)=1$, we get $1=\frac{1}{2}f(c)f'(-c) + \frac{1}{2}f(-c)f'(c)$

Answer 1

Fix a point $a$. For all $h\ne0$, $$f(a+h)-f(a)=f(a)(f(h)-1)$$ Divide by $h$, and use $f$ differentiable at $0$ : $$\frac{f(a+h)-f(a)}{h}=f(a)\frac{f(h)-f(0)}{h}\xrightarrow[h\to0]{}f(a)\times 1$$ Which proves that $f'(a)$ exists and $f'(a)=f(a)$.

Answer 2

You can't use $f'(x)$ before having proved it exists.

Consider instead that $$ f(x+h)-f(x)=f(x)f(h)-f(x)= f(x)\bigl(f(h)-1\bigr)=f(x)\bigl(f(0+h)-f(0)\bigr) $$