Question

Here's something that's bothered me for a while, and that is finding the derivative of x^x.

Answer 1

So we can do it two ways, first the way I know works:
\[y=x^x\]\[lny=xlnx\]\[\frac{ y' }{ y }=lnx+1\]\[y'=x^x(lnx+1)\]
Now the alternate way is to do this, but I don't know why the terms add together.

\[y=x^x\]If we consider x as though it were constant:\[y'=x^xlnx\] And if we do the power rule:\[y'=x*x^{x-1}=x^x\]Add together and simplify:\[y'=x^x(lnx+1)\]

So why does this second method work? It seems to work all the time, unless someone can show me a counterexample, which I would really appreciate.

Answer 2

if you consider x as a constant then y' of x^x will be 0....

Answer 3

The second method works... because it can.

Answer 4

derivative of a constant is 0

Answer 5

No, the derivative of a non-zero constant is the constant raised to zero power minus 1.
@nubeer

Answer 6

constant raised to constant is a constant...

Answer 7

derivative of constant is 0

Answer 8

Ther seems to be an error in your steps in the second method, but if I get your meaning, the two are equivalent, in one you are using a logarithm and the other is more like implicit differentiation.

Answer 9

lets suppose ur constand x=2 or any other number just plug in and take the derivative u can see urself what will u get.

Answer 10

seems? there's definitely an error...

Answer 11

There definitely seems to be an error, yes.

Answer 12

"seeming" does not imply indefinitude...

Answer 13

that was oxymoronic...definitely seems...

Answer 14

Check a dictionary for the usages of 'seem' @lgbasallote

Answer 15

i *am* the dictionary

Answer 16

Very well. Thanks for your input. ;-)

Answer 17

Sorry, perhaps I should clarify because it's missing the point. The "x being constant" just means the method in which I took the derivative as if I was taking the derivative of b^x.

Answer 18

\[D[b^x]=b^x~ln(b)\]\[D[x^b]=b~x^{b-1}\]
\[b^x~ln(b)+b~x^{b-1}\]
\[b^x(ln(b)+b^{1-x}x^{b-1})\]
\[x^x(ln(x)+x^{1-x}x^{x-1})\]
\[x^x(ln(x)+x^{1-x+x-1})\]
\[x^x(ln(x)+1)~:~x\ne0\]

might be sheer coincidence

Answer 19

I think we can prove that it's not sheer coincidence by substituting other values into the formulas and finding that indeed, you do get the right answers. It's just strange that it's the sum of the two methods. Fun right?