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+h1)h+...x^x)/h
This is also equal to lim a>0 lim h>0 (x^(x+a)+(x+h)x^(x+h1)h...x^x)/h.
Evaluating for a=0 you get lim h>0 (x^x+(x+h)x^(x+h1)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+h1)... which when evaluated for h=0 gives us: x(x^(x1)). 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
