Derivative of General Exponential Function/Proof 2

Theorem

Let $a \in \R: a > 0$.

Let $a^x$ be $a$ to the power of $x$.

Then:

$\map {\dfrac \d {\d x} } {a^x} = a^x \ln a$

$\ds \lim_{h \mathop \to 0} \frac {a^{x + h} - a^x} h$

$=$

$\ds a^x \lim_{h \mathop \to 0} \frac {a^h - 1} h$

$\ds $

$=$

$\ds a^x \lim_{h \mathop \to 0} \frac {\map \exp {h \ln a} - 1} h$

$\ds $

$=$

$\ds a^x \lim_{h \mathop \to 0} \paren {\frac {\map \exp {h \ln a} - 1} {h \ln a} } \paren {\frac {h \ln a} h}$

$\ds $

$=$

$\ds a^x \lim_{h \mathop \to 0} \paren {\frac {h \ln a} h}$

$\ds $

$=$

$\ds a^x \ln a$

$\blacksquare$