I don't get how the derivative of the exponential function $\displaystyle a^x$ is derived.

$\displaystyle \frac {dy}{dx}=\lim_{h \to 0} \frac {a^{x+h}-a^x}{h}$

$\displaystyle =\lim_{h \to 0} \frac {a^x * a^h-a^x}{h}$

$\displaystyle =a^x \lim_{h \to 0} \frac {a^h-1}{h}$

$\displaystyle =a^x \lim_{h \to 0} \frac {a^h-a^0}{h}$

This is the point where I start to not get.

$\displaystyle =a^xf'(0)$

(why does $\displaystyle \frac {a^h-a^0}{h}$ = $\displaystyle f'(0)$)

The value of $\displaystyle f'(0)$ is the derivative of $\displaystyle y=f(x)=a^x$ at $\displaystyle x=0$

Assuming that $\displaystyle e$ is the value of $\displaystyle a$ at which the slope of the tangent line at $\displaystyle f(0)$ is 1. (Why assume e?)

I get the rest.

I got a question for the differentation method of $\displaystyle \ln x$ as well.

Since $\displaystyle \lim_{t \to \infty} \bigg ( 1+\frac {1}{t}\bigg )^t=e$

$\displaystyle \lim_{k \to 0}\ln(1+k)^\frac {1}{k}=1$ (how?)