Proving Derivative of $e^x$ Is there a way to prove that the function $e^x$ is its own derivative given only that $$e=\lim_{n\rightarrow\infty}(1+\frac{1}{n})^{n}$$?  I can only seem to find proofs where $e$ being the base such that $e^{x}$ passes through $(0,1)$ with a derivative of one is a given, or that utilize logarithmic differentiation. 
 A: Actually, the simplest way to find the derivative of $e^x$ is to first define $ln(x)= \int_1^x \frac{1}{t}dt$.  From that it is easy to prove the usual properties of $ln(x)$, that $ln(xy)= ln(x)+ ln(y)$ and $ln(x^a)= aln(x)$. And, of course, that $\frac{dln(x)}{dx}= \frac{1}{x}$ follows from the "fundamental theorem of Calculus". 
Then define $e^x$ to be the inverse function to ln(x).  Then it is immediate that, with $y= e^x$, $\frac{dy}{dx}= \frac{1}{\frac{dx}{dy}}= \frac{1}{\frac{1}{y}}= y= e^x$.
A: Thoughts:  Can we prove $f(x) = \lim_{n\rightarrow \infty} f_n(x)$ then $f'(x) = \lim_{n\rightarrow \infty} f_n'(x)$ when $f_n'$  and $\lim f_n'(x)$ exist?
[I leave this as an exercise to the reader....]
Proving $e^x = \lim_{n\to\infty}(1 + x/n)^{n}$ is relatively easy as $\lim_{n\to\infty}(1 + x/n)^{n}=\lim_{xn\to\infty}(1 + x/n)^{n}=\lim_{n\to\infty}(1 + 1/n)^{xn}= (\lim_{n\to\infty}(1 + 1/n)^{n})^x = e^x$.
So if we can do the first (which might be hard) we have
$(e^x)' = \lim  n(1 + x/n)^{n-1}*\frac 1n = \lim (1 + x/n)^{n-1} = (\lim (1+1/n)^{n-1})^x$.
and $e = \lim (1+1/n)^n = \lim(1+ 1/n)^{n-1}(1+1/n) = \lim(1+1/n)^{n-1}\lim (1+1/n) = \lim(1+1/n)^{n-1}*1=\lim(1+1/n)^{n-1}$
So $(e^x)' = (\lim (1+1/n)^{n-1})^x=e^x$.
