If $f'(0)$ exists then show that $f$ is differentiable everywhere. Suppose $f'(0)$ exists and that $f(x+y)=f(x)f(y)$ for all $x\in\mathbb{R}$. Show that $f$ is differentiable for all $x\in\mathbb{R}$. 
I'm not exactly sure how to go about this. I'm not looking for a full solution, but rather i'm looking for a hint to start this off. All I know is that $$f'(0) = \lim_{x \rightarrow 0} \frac{f(x)-f(x_0)}{x-x_0}$$.
EDIT: It appears that I was using a nonoptimal definition of the derivative for this exercise. Thank you for the responses!  
 A: Have a look at the difference quotient
$$\frac{f(x + \Delta) - f(x)}{\Delta} = \frac{f(x)f(\Delta) - f(x)}{\Delta}$$
See what to do now?
A: You don't really need that $f$ is differentiable at $0$ -- being continuous at $0$ is sufficient for the conclusion.
Assume $f$ is not the everywhere zero function (which is clearly differentiable everywhere).
From the functional equation $f(x+y)=f(x)f(y)$ you can prove that $f(0)=1$, then $f(x)\ne 0$ everywhere, then $f(x)>0$ everywhere, and finally $f(x)=(f(1))^x$ for all $x\in\mathbb Q$.
Then, since $f$ is continuous at $0$, it will be continuous anywhere, and therefore it will equal $(f(1))^x$ for all $x$. And $(f(1))^x$ is differentiable everywhere.
A: Hint
Write the derivative as
$$f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}=\lim_{h \to 0} \frac{f(x)f(h)-f(x)}{h}.$$
A: $f'(x)$ is defined to be $\lim_{y \to 0}\dfrac{f(x+y) - f(x)}{y}$ and
$$\lim_{y \to 0}\dfrac{f(x+y) - f(x)}{y} = \lim_{y \to 0}\dfrac{f(x)f(y) - f(x)}{y} = f(x)\lim_{y \to 0}\dfrac{f(y) - 1}{y} = f(x)f'(0)$$
since $f(0)= 1$ if $f$ is not identically $0$
If $f$ is identically $0$, $f'(x) = f(x)f'(0)$ still holds.
A: For this function not to be constant and equal to $0$, $f(0)=1$. Fix an $x\in \mathbb{R}$. For $f$ to be differentiable at $x$, this limit must exist:
$$
\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{f(x)(f(h)-1)}{h}=f(x)\lim_{h\to 0}\frac{f(h)-f(0)}{h}
$$
And this last limit exists, given that the function is differentiable at $0$.
