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The function $f(x)=x^x$ is only defined for negative values of $x$ if $x$ is a rational number $a/b$, where $b$ is odd. Initially, I thought this meant that the limit $$ f'(a)=\lim_{h\to0}\frac{(a+h)^{a+h}-a^a}{h} $$ cannot exist if $a$ is a negative number, since every neighbourhood of $a$ would contain values for which $x^x$ does not make any sense at all. However, this post—about $\lim_{x\to0}\frac{\sin(1/x)}{\sin(1/x)}$—suggests that provided every neighbourhood of $a$ contains values for which $x^x$ is defined, there is no issue with taking limits at the point.

So if $a$ is negative number that is in the domain of $f$, then does $f'(a)$ exist? And if so, is there a general formula for $f'(a)$?

Joe
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  • @Ian: The definition I had in mind was that if $b$ is a rational number $p/q$, and $q$ is odd, then $a^b$ should mean $\left(\sqrt[q]{a}\right)^p$. But I would be very interested in hearing about the extension to all negative real values of $x$. I'm also curious about whether the limit $$\lim_{h\to0}\frac{(a+h)^{a+h}-a^a}{h} , ,$$exists, even if we decide not to extend the definition of exponentiation to all negative real values. – Joe Mar 25 '21 at 17:25
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    The post in your link is about $\dfrac{\sin\frac1x}{\sin\frac1x}$, not about $\dfrac{\sin x}{\sin x}$. The difference is crucial! – TonyK Mar 25 '21 at 18:00
  • @TonyK: Thanks for correcting my error. The difference is indeed crucial, because the former is a very simple example of a function that is undefined for arbitrarily small values of $x$. – Joe Mar 25 '21 at 18:01
  • @TonyK: On the other hand, $\frac{\sin x}{\sin x}$ is a good example of a function which is undefined for arbitrarily large values of $x$, meaning that it was the basis for another very good question. – Joe Mar 25 '21 at 18:04
  • You might be interested in https://math.stackexchange.com/questions/394110/can-the-graph-of-xx-have-a-real-valued-plot-below-zero – Michael Hoppe Mar 25 '21 at 20:23

3 Answers3

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In the usual sense of rational exponentiation, if $x=\frac{a}{b}$ where:

  • $b$ is odd
  • $a$ and $b$ are coprime
  • $x<0$

then $x^x=(-1)^a (-x)^x$. Given any $x<0$ there exist numbers of this form with $a$ even arbitrarily close to it, as well as numbers with $a$ odd arbitrarily close to it, so there is no continuous extension. Moreover we can't take limits or derivatives even at points in the domain, because limits along sequences with $a$ even and limits along sequences with $a$ odd will yield different values.

Ian
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    The sign is not consistently negative, though. If $a$ is even, then $x^x$ should equal $+(-x)^x$, so this extension cannot be extended continuously to negative values of $x$, since any interval contains rational numbers of both even and odd numerator in lowest terms. – Rivers McForge Mar 25 '21 at 17:25
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    @RiversMcForge Thanks for the correction, that was a very careless mistake on my part. I have fixed it now. – Ian Mar 25 '21 at 17:28
  • Does this plot look right: Wolfram Alpha – Jeppe Stig Nielsen Mar 25 '21 at 18:04
  • @JeppeStigNielsen If I understand what you did correctly, you made the same mistake I did in the first version of the answer. The point is that the sign bounces around wildly on arbitrarily small length scales, since it is controlled by the parity of $a$. – Ian Mar 25 '21 at 18:18
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Complex values allowed...
Define $$ x^x = \exp\big(x\log x\big) $$ using the principal value of $\log x$. Undefined at $x=0$.
Then the derivative is $$ \frac{d}{dx}\; x^x = x^x (1+\log x) $$ Here is the real part of the derivative
RE
It is continuous, except at $x=0$, where it goes to $-\infty$.

Here is the imaginary part of the derivative
IM
It is continuous, except at $x=0$, where it jumps from $\pi$ to $0$.

GEdgar
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  • Isn't the negative real axis the location of the branch cut for $\operatorname{Log}$? – Ian Mar 25 '21 at 22:13
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Differentiability is typically only defined for functions defined on an open subset of $\Bbb R$. So although you could make the case that the limit exists because the domain of the function is $\{x\in\Bbb R: x>0 \text{ or }x=a/b\text{ for odd }b\}$, this is not enough to make the function differentiable at any $x<0$.

TonyK
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  • The asker is aware of this. They are asking if there is an extension of $f$ which we could differentiate. – Duncan Ramage Mar 25 '21 at 18:13
  • @DuncanRamage: No, that is not what they are asking. Perhaps that is a question that you might like to ask, in which case feel free to post your own question. – TonyK Mar 25 '21 at 18:16