I've been trying to prove to myself that $\begin{equation} \frac {d}{dx} \end{equation} a^x = a^xln(a)$, and have come across a snag with figuring out how to prove $\begin{equation}ln(a) = \lim \limits_{h \to 0} \frac{a^h-1}{h} \end{equation}$ without using any prior calculus knowledge, save for how to derive basic derivitive rules for polynomials and trig functions.
What I've been able to derive so far is that $\begin{equation} \lim \limits_{h \to 0} \frac{a^h-1}{h} \end{equation}$ is the slope of $\begin{equation} a^x \end{equation}$ at zero, which I can approximate and anylize to figure out that if if $\begin{equation} d(x) = \lim \limits_{h \to 0} \frac{x^h-1}{h} \end{equation}$:
$\begin{equation} d(x) \end{equation}$ must be defined and continuous for $\begin{equation} (0, \infty) \end{equation}$
$\begin{equation} d(x) \end{equation}$ shares a reversed domain and range with $\begin{equation} a^x \end{equation}$
$\begin{equation} d'(x) > 0 \end{equation}$
$\begin{equation} d''(x) < 0 \end{equation}$
$\begin{equation} d(1) = 0 \end{equation}$
Out of all of these, the second seems to be the one with the most promise, but I simply can't find anything more than a basic explanation of inverse functions online or hyper dense mathematics talk that ends up just assuming that $\begin{equation} \lim \limits_{h \to 0} \frac{a^h-1}{h} = ln(a) \end{equation}$, which isn't helpful.