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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 $|x^q-L|<\epsilon$.

Keshav Srinivasan
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  • Binomial theorem. – copper.hat Jul 31 '18 at 15:29
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    @copper.hat Binomial theorem only allows you to derive this for $r\in \Bbb{N}$ – Michal Dvořák Jul 31 '18 at 15:30
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    @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 he meant something that's usually called Binomial series – Jakobian Jul 31 '18 at 15:32
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    @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
  • Can you define irrational exponents without $e$? If yes, then you can prove the power rule using that definition. – Paramanand Singh Jul 31 '18 at 15:33
  • @KeshavSrinivasan You didn't mention logarithms, you only mentioned $e$. Edit your question then. – Michal Dvořák Jul 31 '18 at 15:35
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    @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
  • Your definition needs to be made precise. For example you can try with $x>1$ and then write $x^r=\sup{x^t:t\in\mathbb {Q}, t<r} $ and then work according to that. – Paramanand Singh Jul 31 '18 at 15:37
  • @MichalDvořák I think my question is clear enough as is. I said in my question that I don’t want to use the definition $x^r=e^{r\ln x}$ and instead want to use the definition involving limits over rational numbers. – Keshav Srinivasan Jul 31 '18 at 15:37
  • Another which is more in line with your definition is to define $x^r=\lim_{n\to\infty} x^{r_n} $ where $r_n$ is a sequence of rationals tending to $r$. – Paramanand Singh Jul 31 '18 at 15:38
  • See the later part (after update) of this answer https://math.stackexchange.com/a/1782225/72031 and you need to prove the corresponding inequalities for irrational exponents and you are done. – Paramanand Singh Jul 31 '18 at 15:44
  • @ParamanandSingh What is imprecise about my definition? The limit of a function $f$ from $Q$ to $R$ where the limit goes to a real number is perfectly well defined. We say that the limit of $f(q)$ as $q$ goes $r$ if for any $\epsilon > 0$ there exists a $\delta > 0$ such that for any rational number $q$ such that $|q-r| <\delta$, we have $|f(q)-L|<\epsilon$. – Keshav Srinivasan Jul 31 '18 at 15:45
  • If you deal with functions from $\mathbb{Q} $ to $\mathbb{R} $ then your limit notation $q\to r$ should have both $q, r$ as rational. – Paramanand Singh Jul 31 '18 at 15:47
  • @ParamanandSingh See my edit. – Keshav Srinivasan Jul 31 '18 at 15:51
  • If you can manage to work with your definition then you can try to prove the inequalities for irrational exponents mentioned in my linked answer with your definition. – Paramanand Singh Jul 31 '18 at 15:58
  • @ParamanandSingh Am I imagining things or at some point in the past did you write a blog post on your personal blog defining $x^r$ as the limit of $x^{r_n}$ where $(r_n)$ is a sequence of rational numbers approaching $r$? – Keshav Srinivasan Nov 14 '19 at 02:12
  • @KeshavSrinivasan: I have written a series of posts on theory of exponential and logarithmic functions long back and these posts are famous on MSE (sorry for bragging). You can see them on my archives page: https://paramanands.blogspot.com/p/archives.html?m=0 – Paramanand Singh Nov 14 '19 at 02:15
  • @KeshavSrinivasan : also see https://math.stackexchange.com/a/1603028/72031 – Paramanand Singh Nov 14 '19 at 02:19

2 Answers2

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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 $(0,\infty)$.

Eric Wofsey
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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.