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 Question from David, a student: So, its David, and I was wondering about the derivative of y=x^x. I have often seen it be shown as x^x(ln(x)+1), but when I did it through limits it turned out differently. Here's what I did: It is commonly know that df(x)/dx of a function is also the limit as h->0 of f(x+h)-f(x)/h. To do this for x^x you have to start with lim h->0 ((x+h)^(x+h)-x^x)/h. The binomial theorem then shows us that this is equal to lim h->0 (x^(x+h)+(x+h)x^(x+h-1)h+...-x^x)/h This is also equal to lim a->0 lim h->0 (x^(x+a)+(x+h)x^(x+h-1)h...-x^x)/h. Evaluating for a=0 you get lim h->0 (x^x+(x+h)x^(x+h-1)h...x^x)/h Seeing as the last 2 terms on the numerator cancel out you can simplify to a numerator with h's is each of the terms, which you can then divide by h to get: lim h->0 (x+h)x^(x+h-1)... which when evaluated for h=0 gives us: x(x^(x-1)). This statement is also equal to x^x. This contradicts the definition of the derivative of x^x that is commonly shown. So, my question is: can you find any flaws in the logic of that procedure? I do not want to be shown how to differentiate x^x implicity because I already know how to do that.

Yes - you are taking limits termwise within an infinite sum, which is not generally valid. The result is that you are differentiating as if you were working with (x+h)a for a fixed natural number a that "happened to equal x", and you get the appropriate result!

It's a cute effort though, David, and the flaw is subtle. I believe the College Math Journal is still running their "Fallacies, Flaws, and Flimflam" column, and I suggest that you submit it to them; I think people would enjoy it. http://www.maa.org/pubs/cmj.html

Good Hunting!
RD

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