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.
 A: 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, $\;h\to 0\iff x=h\log a\to 0\;$
A: I must appreciate effort put by OP to define the exponential function $a^{x}$ for $a > 0$ by extending the algebraical definition when $x$ is rational to the case where $x$ is irrational by using continuity argument. While there is nothing wrong with this approach it turns out to be one of the difficult routes to a theory of logarithmic and exponential function.
Now back to the question at hand. Differentiation by first principle of $f(x) = a^{x}$ involves the evaluation of limit $$L(a) = \lim_{h \to 0}\frac{a^{h} - 1}{h}$$ The challenge here is not to find $L(a)$ but to prove that this limit exists. Clearly the limit wont exist unless we have $\lim_{h \to 0}a^{h} = 1$. So as a part of definition of $a^{x}$ we must ensure that we have established $\lim_{h \to 0}a^{h} = 1$.
Note that if $a = 1$ then the limit is $0$ trivially. So let $a \neq 1$ and then there are two cases $a > 1$ and $0 < a < 1$. Clearly by putting $a = 1/b$ we can see that $L(1/a) = L(b) = -L(a)$ (note while proving this we will need $\lim_{h \to 0}a^{h} = 1$) and hence it is sufficient to consider the case $a > 1$.
Now inequalities come to the rescue. From this answer we have $$\frac{a^{r} - 1}{r} > \frac{a^{s} -1 }{s}$$ where $r, s$ are positive rationals and $r > s$. Note that by continuity arguments the inequality can be extended to positive irrational values of $r, s$ with $r > s$ but then the inequality weakens to $\geq$. There are ways to make this inequality strict for irrationals $r, s$ but we won't need the strict version here. Clearly from the above we can see that the function $g(h) = (a^{h} - 1)/h$ is an increasing function of $h$ for $h > 0$. Clearly since $a > 1$ it follows that $g(h) > 0$ for all $h > 0$. Now as $h \to 0^{+}$ the function $g(h)$ decreases but is bounded below by $0$ hence tends to a limit $L(a)$.
If $h \to 0^{-}$ then we can put $h = -k$ and see that $$\lim_{h \to 0^{-}}\frac{a^{h} - 1}{h} = \lim_{k \to 0^{+}}\frac{1 - a^{k}}{-ka^{k}} = \lim_{k \to 0^{+}}\frac{a^{k} - 1}{k} = L(a)$$ It now follows that $g(h)$ tends to a limit as $h \to 0$ which we have denoted by $L(a)$.
By further careful considerations it can be shown that $a > 1$ implies that $L(a) > 0$ and since $L(1/a) = -L(a)$ we have $L(a) < 0$ if $0 < a < 1$. It can be further established using inequalities that $L(a) $ is a strictly increasing function of $a$ for $a > 0$. This function $L(a)$ is traditionally written as $\log a$. Simple properties like $\log(ab) = \log a + \log b$ are provable very easily using this definition. Using this we also get $\log(a^{n}) = n\log a$ for any integer $n$ which shows that range of this $\log $ function is $(-\infty, \infty)$.
It is now a simple matter to show that $(a^{x})' = a^{x}\log a$. Next we can define $e$ by $\log e = 1$ and then $(e^{x})' = e^{x}$ and we can prove that $e^{\log a} = a$ for $a > 0$ and $\log (e^{a}) = a$ for all $a$. Thus $\log x$ and $e^{x}$ are inverses and $(\log x)' = 1/x$ by rule for differentiation of inverse functions. I hope you can proceed along these lines to develop full theory of exponential and logarithmic functions.
A: Well, historically that is just the definition of the natural logarithm, so define
$$l(a)=\lim_{n\to\infty}n(\sqrt[n]{a}-1)$$
and conclude that this function has the usual properties of a logarithm, like $l(ab)=l(a)+l(b)$.
A: 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:
$$
\frac{d}{dx}a^x=a^x\ln(a)
$$
