I am reading R. Shankar's Basic Training in Mathematics. In the first chapter, the author is attempting to give meaning to what $a^x$ means for any (potentially irrational) x.
As the very first step, he presents the following method of determining the derivative of $a^x$ (before either $a^x$ or $ln$ have been defined):
[1] $\Delta a^x$ = $a^{(x + \Delta x)} - a^x$
[2] = $a^x(a^{\Delta x} - 1)$
[3] = $a^x(1 + \ln(a)\Delta x + ... - 1)$
[4] $\frac{da^x}{dx}$ = $a^x\ln(a)$
He justifies step [3] by saying that it is obvious that $a^{\Delta x}$ is very close to 1, and that:
The deviation from 1 has a term linear in $\Delta x$, with a coefficient that depends on $a$, and we call it the function $\ln(a)$...
How is he justified in asserting that, for any choice of $a$, the deviation from $1$ must be represented by a term linear in $\Delta x$?
I can see that, if there were any higher-order terms $a_2(\Delta x)^2$, $a_3(\Delta x)^3$, ..., they would all disappear as $\Delta x \rightarrow 0$ in the derivative, but how do we know that a suitable linear function even exists?
It is a little sketchy.
$\frac {d}{dx} a^x = \lim_\limits{h\to 0} \frac {a^{x+h} - a^x}{h} = a^x \lim_\limits{h\to 0} \frac {a^{h} - 1}{h}$
And then with a wave of the hand.... we assume that this limit exists. And lets call it, "$\ln a.$"
However, it has not been yet been demonstrated that this function of $a$ has the properties of a logarithm, or what the base of this logarithm might be.