Derivative of $a^x$ from first principles 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.

An interesting way to prove it is the following:
Assuming you can prove $\frac{d}{dx}\ln(x)=\frac{1}{x}$ you can write:
$$
x=\ln(x)\Rightarrow\ln(a^x)=x\ln(a).
$$
Thus:
$$
\frac{d}{dx}\ln(a^x)=\frac{d}{dx}x\ln(a)=\ln(a),
$$
But for the chain rule, $\frac{d}{dy}\ln(y)=\frac{1}{y}\frac{d}{dy}y$, hence:
$$
\frac{d}{dx}\ln(a^x)=\frac{1}{a^x}\frac{d}{dx}a^x.
$$
Therefore:
$$
\frac{1}{a^x}\frac{d}{dx}a^x=\ln(a).
$$
Finally:
$$
\frac{d}{dx}a^x=a^x\ln(a)
$$
