Prove that the derivative of exponential is itself without using derivative It's a little problem that I found interesting . To solve it without using differentiation we solve the following equation :
Let $x,y$ be real numbers then solve :
$$e^x=\frac{e^x-e^y}{x-y}$$
We multiply by $(x-y)$ it gives :
$$e^x(x-y)=e^x-e^y$$
Or:
$$e^x(x-y-1)=-e^y$$
We make the following substitution $u=x-y-1$
We have :
$$e^{u+y+1}(u)=-e^y$$
Or 
$$ue^u=-e^{-1}$$
We introduce the product log function 
it gives :
$$W(-e^{-1})=u$$
Or :
$$u=-1$$
So we have $x=y$ and conclude that the derivative of exponential is itself .
My question :
Have you a similar method wich doesn't use derivative .
Thanks a lot for your interest
 A: Perhaps, you will  not like the following solution because I will use differentiation (but not in the case of $e^x$.)
So I try to find a function whose derivative is itself. I cannot be this smart but assume that I try the following. What about $$f_1(x)=1+x?$$ The derivative is $1$. So my first choice is wrong. But, "I have an idea". Let 
$$f_2(x)=1+x+\frac12 x^2.$$
The derivative is wrong again but I have a more general idea now. What about $$f_n(x)=1+x+\frac12x^2+\frac1{3!}x^3+\dots+\frac1{n!}x^n.$$
The derivative is still wrong but promising:
$$f_n(x)=1+x+\frac12x^2+\frac1{3!}x^3+\dots+\frac1{(n-1)!}x^{n-1}.$$
So, I expect that the function which I define the following way will be OK:
$$f_n(x)=1+x+\frac12x^2+\frac1{3!}x^3+\dots+\frac1{n!}x^{n}+\dots$$
Let's call the limit of the series above (if it exists):
$$e^x.$$
This "imaginary" function is, hopefully a function whose derivative is itself. 
A: If you mean to prove that $(e^x)'=e^x$ without using the definition of the derivative (directly), then you could do it as follows:
Let $u=\ln(x)$, then $$e^u=x$$
differentiating both sides, and using the chain rule, we get 
$$(e^u)'u'=1\\(e^u)'\frac{1}{x}=1\\(e^u)'=x\\\boxed{(e^u)'=e^u}$$
