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How to prove that $\frac{d}{dx} x^n = nx^{n-1}$ for every $n>0$ (possibly fractional)?

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It was already shown that $\frac{d}{dx} x^n = nx^{n-1}$ for positive integer $n$. My friend told me that the general case is pretty tricky because in order to prove it, I firstly need to be able to define $x^n$ in each case (rational and irrational real numbers). He made an example using the exponential function, $$e^x = 1 + x + \frac12 x^2 + \frac16 x^3+\dots $$ However, I have not yet covered Taylor series or Maclaurin series.

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Lev Zerkal
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    http://math.stackexchange.com/questions/70979/mathematics-engineering-how-do-you-prove-the-power-rule – lemon Aug 02 '14 at 22:53
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    This is all sorts of confusing. Is your issue with power rule or $e^x$? It's not clear at all. – Cameron Williams Aug 02 '14 at 22:53
  • Please tell us what you mean by "the power rule", and what the power rule has to do with $e^{x\log(21/13)}$, and whether you know what a Maclaurin series is. There is no way to answer your question, when you haven't stated it clearly enough for people to figure out what your question is. – Gerry Myerson Aug 02 '14 at 23:27
  • "It was shown that d/dx x^n = nx^n-1. But this was only for positive integers n. Assuming that n>0, let n be any real number. Prove that d/dx x^n = nx^n-1." My friend told me that this is pretty tricky because in order to prove it, I firstly need to be able to define each case (rational and irrational real numbers). He made an example by showing that e^x = 1 + x + 1/2x^2 + 1/6x^3.... I have not yet covered Taylor series or Maclaurin series – Lev Zerkal Aug 02 '14 at 23:32
  • In any such argument, you need to define what $x^a$ means if $a$ is not rational. Either it is the unique continuous extension of $x^a$ (which is easy to define for rational $a$), or it is $\exp(a \ln(x))$. Showing that these definitions are equivalent is not a completely trivial matter. Proving the power rule from either of them is straightforward, although it is easier from the second definition. – Ian Aug 02 '14 at 23:40
  • I think, Lev, what you have put in the comment you should edit into the question. People shouldn't have to track through the comments to understand what the question is. – Gerry Myerson Aug 03 '14 at 07:13

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We want to find the derivative of $x^a$, for positive $x$. Note that $x^a=\exp(a\ln x)$. (For general $a$, that is precisely how $x^a$ is ordinarily defined.).

Differentiate, using the Chain Rule.

We get $\frac{a}{x}\exp(a\ln x)$. This is $\frac{a}{x}x^a$, which is $ax^{a-1}$. That is the rule that we already knew held when $a$ is a positive integer.

André Nicolas
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  • Your definition of $x^a$ does not seem to help Lev. Either you are defining it as the irrational number raised to a (possibly) irrational number, in which case this suffers from the same problem as $x^a$, or you are defining it using a Taylor series which Lev does not know. Am I missing something? – RghtHndSd Aug 14 '14 at 02:23
  • I am assuming that the exponential function has been defined in the calculus course. Since OP has not reached power series, it may have been done in the way common in North American calculus books. Define $\ln x$ using an integral, and let $\exp$ be its inverse function. Then we can calculate the derivative of $\exp$, and the answer above becomes available. I agree that if $\exp$ has not been defined, then the answer is not helpful. One could define irrational powers as a limit of rational powers. For good reason it is not done that way: verifying the basic properties is too painful. – André Nicolas Aug 14 '14 at 02:36
  • Ah, I knew I was missing something! – RghtHndSd Aug 14 '14 at 02:44