Given the power rule $D(x^n) = n \cdot x^{n-1}$:

My calculus book proves that for irrational $n$, it holds for $x > 0$. Wikipedia’s aricle on the Power Rule is confusing me because it makes no mention of this limitation. And given a problem such as $D(x^π)$, my book does not give an answer with the limit of $x>0$. (Is that because $x^π$ is itself not defined if $x\not>0$, so it is “by definition” limited?)

So is there some extension I don’t know of that extends it to $x\leq0$?


2 Answers 2


It depends. One possible definition of $x^\alpha$ for $\alpha $ a real number is

$$ x^\alpha = e^{\alpha \cdot \ln x} \ . $$

Of course, if you're just talking about real functions, then this expression is not defined for $x\leq 0$. But if by $\ln$ you mean the complex logarithm, then it makes perfectly sense and you can check that the rule for deriving it is still true. Namely,

$$ D(x^\alpha) = \alpha \cdot x^{\alpha - 1} \ . $$

  • $\begingroup$ I'm just talking about real functions. Just to clarify the second half of my question: is my calculus book then mistaken by $x^π$? Or is this because like you said that the expression is not defined for $x\leq0$? Is there no need to expressly state it then? $\endgroup$
    – Jeff
    Oct 20, 2013 at 19:11
  • $\begingroup$ Also, out of interest, doesn't this definition preclude even $(-5)^2$? Or is it defined piece-wise, with negative x being defined differently? $\endgroup$
    – Jeff
    Oct 20, 2013 at 19:15
  • $\begingroup$ Then the logarithm you see up there is not defined for $x\leq 0$. $\endgroup$ Oct 20, 2013 at 19:15
  • $\begingroup$ I'm also curious about what you mean by a "real" function vs. a "complex function." It's typically a function $f: \mathbb{R} \to \mathbb{R}$, but would you call the function $f: \mathbb{R} \to \mathbb{C}$ given by $f(x) = \ln{x}$ a complex function? $\endgroup$
    – Ayesha
    Mar 8, 2014 at 17:27
  • $\begingroup$ Nope. I would call $\mathrm{ln} : \mathbb{C}\setminus \{ 0\} \longrightarrow \mathbb{C}$ a complex function. $\endgroup$ Mar 8, 2014 at 19:41

For negative $x$, there is no way to define a real $x^\alpha$ if $\alpha$ is irrational. The (complete, as continuous as possible) definition of powers is that $x^\alpha = \mathrm{e}^{\alpha \ln x}$, when $x$ is negative there is no $\ln x$ in the reals.


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