I've been trying to think for the past few days how one could differentiate $a^x$ based on the definition that $a^n$ is repeated multiplication, $a^{n/m}=(\sqrt[m]a)^n$, and $a^x$ is the completion of the above function by continuity.
With a bit of algebra, the problem quickly reduces to finding the derivative at $0$:
$$\lim_{h\to0}\frac{a^h-1}{h}\tag{1}$$
And that limit's really got me stumped. Since you'll obviously have to use the definition in some way, I thought I'd replace $h$ with $\frac{1}{n}$ and use the $n$-th root definition:
$$\lim_{n\to\infty}n(\sqrt[n]a-1)\tag{2}$$
Of course, just because $(2)$ exists doesn't automatically imply $(1)$ exists, but it might be a first step. Even $(2)$ has me stumped, though.
Best Answer
What about using $\;a^x=e^{x\log a}\;?$ Then
$$\frac{a^h-1}h=\frac{e^{h\log a}-1}h\;\;\stackrel{\text{subst.}\;h\log a\to x}=\;\;\frac{e^x-1}{\frac x{\log a}}\xrightarrow[x\to 0]{}\log a$$
since, in the above substitution, $\;h\to 0\iff x=h\log a\to 0\;$