TITLE: Differentiate $2^x$ from first principles.
QUESTION [5 upvotes]: I am trying to differentiate $2^x$ from first principles. This is what I have so far:
\begin{align}
f'(x) &= \underset{h\rightarrow 0}{\textrm{lim}} \frac{f(x+h) - f(x)}{h}\\
\implies \frac{d2^x}{dx} &= \underset{h\rightarrow 0}{\textrm{lim}}\frac{2^{x+h}-2^x}{h}\\
&= \underset{h\rightarrow 0}{\textrm{lim}}\frac{2^x(2^h -1)}{h}
\end{align}
From that point on, as the limit is of type 0/0, I was thinking of using L'Hôpital's rule, but this gives
\begin{equation}
\frac{d2^x}{dx} = 2^x\frac{d2^h}{dh}\bigg\rvert_{h=0}.
\end{equation}
Not sure how to go from there.

REPLY [7 votes]: Here, we present a way forward the uses pre-calculus tools only.  To that end, we begin with a primer.

PRIMER:
In THIS ANSWER I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities
$$\bbox[5px,border:2px solid #C0A000]{1+x\le e^x\le \frac{1}{1-x}}\tag 1$$
for $x<1$.


Note the $2^h=e^{h\log(2)}$.  Applying $(1)$ reveals
$$1+x\le e^x\le \frac{1}{1-x}$$
for $x<1$.
Then, we can write
$$\log(2)\le \frac{e^{h\log(2)}-1}{h}\le \frac{\log(2)}{1-h\log(2)}$$
whence applying the squeeze theorem yields the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{h\to 0}\frac{2^h-1}{h}=\log(2)}$$