# Differentiate $2^x$ from first principles.

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.

• How do you define $2^x$? See also this question. – Batman Dec 14 '16 at 14:17
• Actually, using l'Hopital rule you have to derive with respect to $h$ the expression in the numerator and denominator so the limit transforms to $\lim_{h\rightarrow 0}\frac{2^{x+h}\log{2}}{1}=2^x\log{2}$ as expected. – Jesús Ros Dec 14 '16 at 14:19

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

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

• is your first inequality or its domain correct? If I plug in x = 2 i get e^2<1/(1-2), which is clearly wrong. – MrYouMath Dec 14 '16 at 15:08
• @MrYouMath If you notice the domain is clearly specified as $x<1$. So, what further questions can I answer. – Mark Viola Dec 14 '16 at 15:11
• @Dr.MV I was talking about your first primer. You have written $x>0$ in this primer. – MrYouMath Dec 14 '16 at 15:22
• @MrYouMath That was a copy-and-paste error. I've edited. – Mark Viola Dec 14 '16 at 15:37
• @Masacroso That was a copy-and-paste error. I've edited. – Mark Viola Dec 14 '16 at 15:37