Solution for exponential function's functional equation by using a definition of derivative let $f(0)=1$ and $f'(0)=1$.
and $f(x+y)=f(x)f(y)$ for $x,y\in R$.
How can I found $f(x)$ by using a definition of derivative?
 A: Let me just turn that last comment into an answer. As you already pointed out in the comments, by the definition of the derivative and using $f(x+y) = f(x)f(y)$, $f(0) = 1$ and $f'(0) = 1$ we have
$$f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h\to 0} \frac{f(x)f(h)-f(x)f(0)}{h} = f(x) \lim_{h\to 0} \frac{f(h)-f(0)}{h} = f(x) f'(0) = f(x)$$
Now you're almost done, as $f'(x) = f(x)$ is one of the essential properties of the exponential function. From $f'(x) = f(x)$ it follows that
$$\frac{f'(x)}{f(x)} = \frac{d}{dx}\left(\ln f(x)\right) = 1$$
Integrating both sides, we get $\ln f(x) = x + C$ for some constant $C$. This implies $f(x) = D e^x$ for some constant $D$ (i.e. $D = e^C$). We know that $f(0) = 1$, so it follows that $D = 1$ and $f(x) = e^x$.
Alternatively, as pointed out by Ragib, after noticing $f'(x) = f(x)$ you could write $f(x) = e^x g(x)$ for some function $g(x)$. Then $f'(x) = e^x g'(x) + f(x) = f(x)$, so $e^x g'(x) = 0$ for all $x$. Since $e^x \neq 0$, it follows that $g'(x) = 0$ and $g(x) = D$ is constant. Using $f(0) = 1$ it then follows that $D = 1$ and $f(x) = e^x$.
