If $f(a+b)=f(a)f(b)$, $f(0)=1$, and $f$ is differentiable at $x=0$, show that $f$ is differentiable for all $x$, and that $f'(x)=f'(0)f(x)$ I know this looks kinda like a duplicate, but I'm going to ask a specific question that I feel wasn't answered in the other question's answers.
So we start with the definition of the derivative. $$\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$$
Then, given $f(a+b)=f(a)+f(b)$ we have $$\lim_{h \to 0}f(x)\frac{f(h)-1}{h}$$ which is $$\lim_{h \to 0}f(x)\frac{f(0+h)-f(0)}{h}=f'(0)f(x)$$
but from here I don't know how to show that f is differentiable for all x because seems as though this only holds for $x≠0$ since the function is explicitly different at $0$. Not sure if this is me just overthinking this problem, but this is really bothering me right now.
 A: Since $f'(0)=\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}$ exists we have $f'(0)=f(0)\cdot\lim_{h\to 0}\frac{f(h)-1}{h}$ and thus $$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=f(x)\cdot\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}=\frac{f'(0)}{f(0)}\cdot f(x)$$
and therefore $f$ is differentiable everywhere with $f'(x)=f'(0)\cdot f(x)$.
A: If what is bothering you is that you seem to be assuming that $f$ is differentiable at $x$ in your proof, you can rewrite the proof as follows. Note that for all $x,h\in\mathbb R$ such that $h\neq0$,
$$
\frac{f(x+h)-f(x)}{h}=\frac{f(x)f(h)-f(x)}{h}=f(x)\cdot \frac{f(h)-1}{h}=f(x)\cdot\frac{f(h)-f(0)}{h} \, .
$$
Now, we can use the "product law" for limits: if $\lim_{x\to a}f(x)$ exists and equals $l$, and $\lim_{x\to a}g(x)$ exists and equals $m$, then $\lim_{x\to a}f(x)g(x)=lm$. By assumption, $f$ is differentiable at $0$, meaning that
$$
f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}
$$
exists. Moreover, $f(x)$ is a constant with respect to $h$, and so $\lim_{h\to0}f(x)=f(x)$. Therefore, using the "product law", we find that
$$
\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}f(x)\cdot\frac{f(h)-f(0)}{h}
$$
exists, and equals $f(x)f'(0)$. That's it.
A: There are a couple of other interesting things about this function:
$$f(a+b)=f(a)f(b)\qquad f(0)=1$$
$$1=f(0)=f(1-1)=f(-1)f(1)$$
and you can apply this in general:
$$1=f(0)=f(x-x)=f(x)f(-x)\Rightarrow f(-x)=\cfrac{1}{f(x)}$$
