Finding the derivative of an exponential function without the chain rule Given $f(x) = a^x$, one can compute the derivative of $f(x)$ using the chain rule quickly by noticing that
$f(x) = a^x = \left(e^{ln(a)}\right)^x$.
But how would you go about computing the derivative of $f(x)$ without the chain rule, with just the $f'(x) = \lim_\limits{h \to 0} \frac{f(x + h) - f(x)}{h}$ or the $f'(x) = \lim_\limits{x \to a} \frac{f(x) - f(a)}{x - a}$ definition of a derivative and the fact that $\lim_\limits{h \to 0} \frac{e^h - 1}{h} = 1$?
$f'(x) = \lim_\limits{h \to 0} \frac{a^{x + h} - a^x}{h} = a^x \cdot \lim_\limits{h \to 0} \frac{a^h - 1}{h} = a^x \cdot \lim_\limits{h \to 0} \frac{\left(e^{ln(a)}\right)^h - 1}{h} = a^x \cdot \lim_\limits{h \to 0} \frac{(e^h)^{ln(a)} - 1}{h}$.
How would you proceed from here?

Here is a way to show $a^x$ has derivative $\ln(a) \cdot a^x$ without using chain rule. It uses the series expansion of $e^x$.
Since $a^x = e^{x \ln(a)}$, we have:
$$
a^x = \sum_{n=0}^\infty \frac{(x\ln(a))^n}{n!}.
$$
We can take the derivative term by term (using your favorite convergence theorem), which yields:
$$
\frac{d}{dx} a^x = \sum_{n=0}^\infty \frac{nx^{n-1} \ln(a)^n}{n!}.
$$
Now we make the very important observation that the first term (when $n=0$) is zero. Hence, we can index starting from $n = 1$ and cancel out the $n$ in the numerator:
$$
\frac{d}{dx} a^x = \sum_{n=1}^\infty \frac{x^{n-1}\ln(a)^n}{(n-1)!} = \ln(a) \cdot \sum^\infty_{n=1} \frac{x^{n-1}\ln(a)^{n-1}}{(n-1)!},
$$
noting we can pull out the factor of $\ln(a)$ from every term now. But the last sum can now be re-indexed,
$$
\sum^\infty_{n=1} \frac{x^{n-1}\ln(a)^{n-1}}{(n-1)!} = \sum_{n=0}^\infty \frac{x^n\ln(a)^n}{n!} = a^x,
$$
so in conclusion we have:
$$
\frac{d}{dx} a^x = \ln(a) \cdot a^x.
$$
