Derivatives of Exponential Functions

Here we derive the formula for the derivative of the exponential function $f(x)=a^x$, namely \begin{align} f'(x_0)= \ln(a)\cdot a^{x_0}\,. \end{align} The derivative of a function $f$ in a point $x_0$ is defined as the limit \begin{align} f'(x_0) = \lim_{h\rightarrow 0} \frac{f(x_0+h) - f(x_0)}{h} \end{align} provided that this limit exists.
We plug $f(x)=a^x$ into this and rearrange: \begin{align} f'(x_0) &= \lim_{h\rightarrow 0} \frac{a^{x_0+h} - a^{x_0}}{h} \\ &= a^{x_0}\cdot \lim_{h\rightarrow 0} \frac{\left(a^h - 1\right)}{h}\,. \end{align} The task is now to determine the limit in the last line. We take its derivative with respect to $a$, \begin{align} \frac{d}{da}\left(\lim_{h\rightarrow 0} \frac{\left(a^h - 1\right)}{h}\right) &= \lim_{h\rightarrow 0} \left(\frac{h\cdot a^{h-1}}{h}\right)\\ &= \frac{1}{a}\,. \end{align} It follows that the limit is the antiderivative \begin{align} \lim_{h\rightarrow 0} \frac{\left(a^h - 1\right)}{h} &= \int \frac{1}{a}\,\mathrm{d}a\\ &= \ln(a) + c\,. \end{align} The derivative of $f$ must then be \begin{align} f'(x_0) &= a^{x_0}\cdot\left(\ln(a)+c\right)\,. \end{align} At $a=1$ the function $f(x)=a^x$ is the constant function $f(x)=1$, and so its derivative must equal zero. Using $\ln(1)=0$ this lets us determine the value of the constant of integration $c$, \begin{align} \ln(1) + c &= 0\\ c &= 0\,. \end{align} The derivative of $f(x)$ is then \begin{align} f'(x_0) &=\ln(a)\cdot a^{x_0}\quad \square \end{align}