Derivative of $a^x$ where $a \in \mathbb{R}^+$ I work in a University Math Tutor Lab, and recently had a student ask for help taking the derivative of $a^x$ where $a \in \mathbb{R}$. I know one can manipulate it into a pretty straightforward Chain Rule,
\begin{align*}
(a^x)'= \left(e^{\ln(a^x)}\right)'=\left(e^{x\ln(a)}\right)'= \ln(a)\,e^{\ln(a^x)}=\ln(a)\,a^x.
\end{align*}
However, I was hoping someone could provide an intuitive explanation to give students why they should expect $\ln(a)$ to be in their derivative rather than just memorizing the process.
(I know one could say that in a similar fashion the derivative of $e^x$ is $\ln(e)\, e^x=e^x$, but I don't see this as giving students a helpful way to remember that the derivative of $a^x$ has $\ln(a)$ in it.)
 A: By the definition of the derivative,
$$\left(a^x\right)'=\lim_{h\to 0}\frac{a^{x+h}-a^x}h=\lim_{h\to 0}\frac{a^h-1}h\,a^x=L(a)\,a^x.$$
From this we see that the derivative of an exponential is an exponential of the same base, times a constant (that depends only on the base).
At this stage, without expanding more on the function $L(a)$, you can state that there exist a constant $e$ such that $L(e)=1$, so that
$$(e^x)'=e^x.$$

For intuition that $L(a)$ is a logarithm, consider
$$\frac{a^h-1}h=l\iff a=(1+hl)^{1/h}=\left(1+\frac ln\right)^n,$$ and you recognize the expression that yields $e^l$ in the limit.
A: If we start with the assumption that $a^x$ is differentiable of the form $(a^x)'=L(a)a^x$ for some multiplicative factor $L(a)$, then we can see that
$$\begin{align}
L(ab)(ab)^x&=((ab)^x)'\\
&=(a^xb^x)'\\
&=(a^x)'b^x+a^x(b^x)'\\
&=L(a)a^xb^x+a^xL(b)b^x\\
&=(L(a)+L(b))(ab)^x
\end{align}$$
so that the function $L$ has the additive property of a logarithm, i.e., $L(ab)=L(a)+L(b)$. Since $(e^x)'=e^x$, we have $L(e)=1$, so $L$ is the natural logarithm, $L(a)=\ln a$.
A: Intuition but no final conclusion:
$[a^x]' = \lim\limits_{h\to 0}\frac {a^{x+h} - a^x}h = \lim\limits_{h\to 0}\frac {a^xa^h - a^x}h=$
$\lim\limits_{h\to 0}a^x\frac {a^h -1}h = a^x \lim\limits_{h\to 0}\frac {a^h-1}h$.... assuming such a limit exists.
It's reasonable (albeit not an article we can take on faith) to expect that $\lim\limits_{h\to 0}\frac {a^h-1}h$ exists (if it didn't then $a^x$ wouldn't be differentiable).  And if someone were to tell us that $\lim\limits_{h\to 0}\frac {a^h-1}h=\ln a$, it shouldn't surprise us.
However to prove $\lim\limits_{h\to 0}\frac {a^h-1}h=\ln a$ will be a matter of definitions and the manner in which $e$ and/or $\ln x$ where introduced and defined.  Which is a (very important) lesson for later.
One thing that should be apparent (although maybe not clear how to prove) that $\lim\limits_{h\to 0}\frac {a^h-1}h$ is continuous and monotonically increasing and unbounded, and that the value is negative for $0 < a < 1$ and positive for $a > 1$ and so there is will be a value of $a$ where  $\lim\limits_{h\to 0}\frac {a^h-1}h=1$ and so $[a^x]' = a^x$.
We could, and many calculus texts do[1], define that value as $e$.  However most texts don't.  If you define $e = \lim (1 + \frac 1n)^n$ then it's probably possible to prove $\lim\limits_{h\to 0}\frac {e^h-1}h=1$.
.....
All this, though, is tap dancing around the issue of how is $a^x$ for $x \not\in \mathbb Q$ in the first place. 
[1] Oh, alright.  Mine did and absolutely no other text I've ever seen since did.  I can't remember how my text defined $a^x$.  As a freshman, it never occurred that there was an issue and that it needed defining. 
