Elementary proof that $2^x$ is derivable

Question

I was wondering if there was an elementary proof, so not using the exponential function, that $2^x$ is derivable. I define the function $f(x) = 2^x$ by $f(a/b) = \sqrt[b]{2^a}$ for a and b integers, and I extend this function by continuity for the rest of the reals. Because we have $\lim_{h\to 0} \frac{2^{x+h} - 2^{x}}{h} = 2^{x} * \lim_{h\to 0} \frac{2^{h} - 1}{h}$, I just need to show that this last limit exists, but I couldn’t do it.

This question is interesting to me, because it could be an elementary way to prove that there is a unique real positive number e, such that $(e^x)’ = e^x$. We see easily that $2^x$ is continuous and strictly increasing, therefore we can define $\log_2(x)$ for $x > 1$. We define $a = \lim_{h\to 0} \frac{2^{h} - 1}{h}$, and we have for all real number $r > 1$, $(r^x)’ = (2^{\log_2(r)*x})’ = \log_2(r)*a*r^x$. If we pose $e = 2^{1/a}$, we have $(e^x)’ = e^x$.

Edit : my bad, I found quite the same question here : Show $\lim\limits_{h\to 0} \frac{(a^h-1)}{h}$ exists without l'Hôpital or even referencing $e$ or natural log, and there’s a pretty neat answer by zhw.

Answer 1

There is a proof based entirely on the methods of differential calculus; see this

Differentiability of Exponential Functions
by Philip M. Anselone and John W. Lee

In that paper you will find the following.

Theorem 1. Let $f (x) = a^x$ with any $a > 1$. Then f is differentiable at $ x = 0$ and $f'(0) > 0$.

Theorem 2. Let $f (x) = a^x$ with any $a > 1$. Then f is differentiable for all $ x$ and $f'(x) = f'(0)a^x$.

The authors continue, justifying the claim that there is one and only one $e > 0$ satisfying

$\tag 1 \frac{d}{dx} e^x = e^x$

They then introduce the natural log function.

Answer 2

In my lecture i give the following proof: Substituting $$t=2^h-1$$ then $$h=\frac{\ln(t+1)}{\ln(2)}$$ and we have $$\frac{t}{\frac{\ln(t+1)}{\ln(2)}}$$ And for $h$ tends to $o$ $t$ tends to zero and we get $$\lim_{t\to 0}\frac{\ln(2)}{\ln(t+1)^{1/t}}=\ln(2)$$ since in the denominator we get $$\lim_{t \to 0}(1+t)^{1/t}=e$$

Answer 3

If I am right, you can extend the rational powers to reals using a Dedekind cut, setting $S=2^R$ (where $R$ is built on rationals) iff the elements of the cut $S$ are the powers of the respective elements of $R$. Continuity over the rationals will follow.

Then

$$(2^x)'=\lim_{h\to 0}\frac{2^{x+h}-2^x}h=2^x\lim_{h\to 0}\frac{2^h-1}h=l_22^x$$

holds if we can show that the limit exists.

Now consider the function $f(n):=n(2^{1/n}-1)$, such that

$$\frac12<f(n)=n\frac{2-1}{2^{(n-1)/n}+2^{(n-2)/n}+\cdots 2^0}<1,$$ and strictly decreasing. Hence the limit exists and is finite.

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A possible next step is to repeat this reasoning for the function $b^x$ and show that

$$2^{1/l_2}=b^{1/l_b}$$ so that we have a "universal" constant $e$ such that $l_e=1$.