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.

  • $\begingroup$ Have you seen this post? $\endgroup$ – Toby Mak Oct 3 '19 at 10:47
  • $\begingroup$ I didn’t see it, but it doesn’t really answer my question. It was a question about the exact derivative of $a^x$ without the log function (which seems impossible), but I just want to prove that $a^x$ is derivable. $\endgroup$ – QuinnLesquimau Oct 3 '19 at 11:20
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    $\begingroup$ @QuinnLesquimau you cannot even define $2^x$ without the exponential and log functions. Please ask yourself what is $2^{\sqrt{2}}$ and how you can define it without using log. Spoiler: you can't multiply the number $2$ "$\sqrt{2}$ times by itself" $\endgroup$ – Glougloubarbaki Oct 3 '19 at 12:20
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    $\begingroup$ @Glougloubarbaki with the definition given above based on extending by continuity (I guess you need to check some uniform continuity condition to make sure this extension exists and is unique), you still get a way to define $2^{\sqrt{2}}$, as a limit of $2^x$ for rational $x$ approaching $\sqrt{2}$. $\endgroup$ – Joppy Oct 3 '19 at 12:29
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    $\begingroup$ @Glougloubarbaki This can be done in a quite elementary way, see here for example. $\endgroup$ – Joppy Oct 3 '19 at 12:55

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.


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 $$\lim_{t \to 0}(1+t)^{1/t}=e$$

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    $\begingroup$ Thank you ! But this proof isn’t really an answer to my question, because it uses the ln function, and I don’t see how to use it without defining the exponential function. $\endgroup$ – QuinnLesquimau Oct 3 '19 at 11:24
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    $\begingroup$ I think your question is impossible to answer without introducing the exponential function. $\endgroup$ – Toby Mak Oct 3 '19 at 11:34

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.


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

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.

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 $l_e=1$.

  • $\begingroup$ It’s an interesting idea, but it’s not obvious to me that the sequence you define is really decreasing. $\endgroup$ – QuinnLesquimau Oct 3 '19 at 12:48
  • $\begingroup$ @QuinnLesquimau: by convexity of $2^x$. You are in fact computing the average value on $[0,1]$, with more and more points. $\endgroup$ – Yves Daoust Oct 3 '19 at 12:57

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