David pose la question : 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 implicitly because I already know how to do that. Robert Dawson lui répond.
Sarah pose la question : Ok, our teacher gave us this riddle, and I cannot for the life of me figure it out. He said that there are three problems with the following proof: Penny Nom lui répond.
