Can you prove the power rule for irrational exponents without invoking $e$?

The power rule states that for any real number $r$,

$$\frac{d}{dx}x^r=rx^{r-1}$$

Now one common way to prove this is to use the definition $x^r=e^{r\ln x}$, where $e^x$ is defined as the inverse function of $\ln x$, which is in turn defined as $\int_1^x\frac{dt}{t}$.

But this puts the cart before the horse, because students typically learn differential calculus before integral calculus. And there is a perfectly good definition of exponentiation of real numbers that does not rely on integral calculus:

$$x^r=\lim_{q\rightarrow r} x^q$$

where $q$ is a variable that ranges over the rational numbers.

So my question is, if we use this definition, and we take it for granted that $\frac{d}{dx}x^q=qx^{q-1}$ holds true for rational numbers (which can be easily proven without invoking $e$), then can we prove the power rule for real exponents without invoking $e$?

EDIT: Here’s a more precise formulation of the definition above. If $r$ is a real number, we say that $x^r = L$ if for any $\epsilon>0$ there exists a $\delta>0$ such that for any rational number $q$ such that $|q-x| < \delta$, we have $|x^q-L|<\epsilon$.

• Binomial theorem. – copper.hat Jul 31 '18 at 15:29
• @copper.hat Binomial theorem only allows you to derive this for $r\in \Bbb{N}$ – Michal Dvořák Jul 31 '18 at 15:30
• @MichalDvořák That’s not a duplicate at all, that’s about proving properties of the natural logarithm function, my question is about deriving the power rule without talking about $e$ and $\ln(x)$ at all. – Keshav Srinivasan Jul 31 '18 at 15:31
• @MichalDvořák: The binomial theorem works for more than natural powers. It avoids using integrals which is what the OP was looking for. – copper.hat Jul 31 '18 at 15:32
• @ParamanandSingh Yes, you can. I gave the definition in my question: $x^r=\lim_{q\rightarrow r} x^q$ where $q$ is a variable that ranges over the rational numbers. I want to know how to use that definition to prove the power rule. – Keshav Srinivasan Jul 31 '18 at 15:35

Yes. In general, if you have a sequence of $C^1$ functions $f_n$ on some interval and functions $f$ and $g$ such that $f_n\to f$ pointwise and $f_n'\to g$ uniformly, then $f'=g$. (Quick sketch of how to prove this without using integration: fix $x$ and $h$ and pick $n$ so that $f_n$ is sufficiently close to $f$ at $x$ and $x+h$ and $f_n'$ is sufficiently close to $g$ uniformly. Then $\frac{f(x+h)-f(x)}{h}$ will be close to $\frac{f_n(x+h)-f_n(x)}{h}$. By the mean value theorem the latter is equal to $f_n'(y)$ for some $y$ between $x+h$ and $x$, and $f_n'(y)$ is close to $g(y)$. Finally, if $h$ is sufficiently small, $g(y)$ will be close to $g(x)$ since $g$ is continuous.)

So, given $r\in\mathbb{R}$, pick a sequence of rational numbers $q_n$ converging to $r$ and let $f_n(x)=x^{q_n}$. Then $f_n(x)$ converges to $f(x)=x^r$ and $f_n'(x)=q_nx^{q_n-1}$ converges to $g(x)=rx^{r-1}$. Moreover, it is not hard to check that these convergences are uniform on any compact subset of $(0,\infty)$. It thus follows that $f'=g$ on $(0,\infty)$.

Hint:

$$x^r:=\lim_{q\to r,\\q\in\mathbb Q}x^q.$$

Then

$$(x^r)'=\lim_{h\to0}\frac{\lim_{q\to r,\\q\in\mathbb Q}((x+h)^q-x^q)}{h}=\lim_{q\to r,\\q\in\mathbb Q}\lim_{h\to0}\frac{((x+h)^q-x^q)}{h}=\lim_{q\to r,\\q\in\mathbb Q}qx^{q-1}=rx^{r-1}.$$

The hard part is to justify the swap of the limits.