There's the rub! It depends, of course, upon how you define e^x.
Many modern Calculus texts first define ln(x)= \int_a^x dx/x and then define e^x as the inverse function so your question never arises. That's my preferred method.
Robert1986 suggests that you define e^x as as power series. That's perfectly reasonable, and again, your question does not arise.
But it is probably most common to define a^x by first defining a^n by "a multiplied by itself n times" then extending to all real numbers by requiring that a^{n+m}= (a^n)(a^m), (a^n)^m= a^{mn}, and continuity. Using that definition, we have
\frac{f(x+h)-f(x)}{h}= \frac{a^{x+h}- a^x}{h}= \frac{a^xa^h- a^x}{h}
= \frac{a^x(a^h- 1)}{h}= a^x\frac{a^h- 1}{h}
So that
\frac{da^x}{dx}= \lim_{h\to 0}\frac{a^{h+1}- a^x}{h}= a^x\lim_{h\to 0}\frac{a^h- 1}{h}
is just C_aa^x where C_a is that limit.
It is easy to show that if 0< a< 1, then C_a is negative, that C_2 is less than 1, that C_3 is greater than 1, and that C_a continuously increases as a increases. That means that there must exist some value of a, between 2 and 3, such that C_a= 1 and we define "e" to be that value of a.
From \lim_{h\to 0}\frac{e^h- 1}{h}= 1 we can say that, for h close to 0, \frac{e^h- 1}{h} is close to 1 so that e^h- 1 is close to h. Then e^h is close to 1+ h and so e is close to (1+ h)^{1/h} which is common limit definition of e: e= \lim_{h\to 0}(1+ h)^{1/h} or, taking n= 1/h, e= \lim_{n\to\infty} (1+ 1/n)^n.
