Differentiability of $a^x$ at $x=0$ I'm sure almost anyone who registered at this forum, knows that $\frac{d(a^x)}{dx} = a^x\ln a \quad $ for $a>0$
But what if you want to actually prove this identity? I'd come along an interesting problem in a standard textbook on calculus where the author asks, "Suppose $f(x)=a^x$ is differentiable at $x=0$, and that $f'(0)=k$ where $k\neq 0$. Prove that $f$ is differentiable at any real number $x$ and that $f'(x)=k\,a^x = k f(x).$", which is no big deal (using definition of differentiation). But I became curious that,
How can one deduce that the differentiation of the function at the $0$ is bounded and equal to $\ln a$?
Any help on approaching this problem would be appreciated in advance.
 A: The derivative of $a^x$ at $x=0$ is
$$ \lim_{h \to 0} \frac{a^h-1}{h}. $$
This is one way of defining $\log{a}$, but of course you then have the problem of defining $a^x$.
If we instead define $\log$, at least initially, as the inverse of the exponential, and then define the general power function as $a^x=e^{x\log{a}}$, the result follows from the chain rule, or using the power series for the exponential.
Alternatively, if one defines the logarithm using an integral, we can obtain the result by exchanging the limit and integral (which is fine because the power functions converge uniformly on the compact interval $[1,a]$):
$$ \frac{a^h-1}{h} = \int_1^a x^{h-1} \, dx \to \int_1^a x^{-1}\, dx = \log{a} $$
as $h \to 0$.
A: This is the chain rule, you note merely that $a^x = e^{x\log a}$. So $u(x) = x\log a$ and then you take derivatives and you get
$${d\over dx}e^u = {d\over du}e^u\cdot {du\over dx}$$
$u$ is a linear function, so the derivative is just the slope, $\log a$ which gives you
$$e^u(\log a) = e^{x\log a}\log a=a^x\log a.$$
But in particular the chain rule tells you this holds at $x=0$ since ${d\over du}e^u$ exists at $u(0)$ and ${d\over dx}(x\log a)$ exists at $0$.
