derivative x^x

[The_Brain]

How do you solve the derivative of x^x? I'm sure it's fairly easy-- I'm just beginning calc though and none of the "forumlas" work.

[StephenPrivitera]

Take the natural log of both sides.
y=xx
lny=xlnx
Differentiate with respect to x.
(1/y)(dy/dx)=lnx+x/x (chain rule, product rule)
dy/dx=y(lnx+1)

You could do the same thing with y=xa, where a is a constant.
lny=alnx
dy/dx=y(a/x)=xa(a/x)=axa-1
a formula I'm sure you know well

[Guybrush Threepwood]

or xx = ex*ln(x) ad continue from here...
(ex*ln(x))' = ex*ln(x) * (x*ln(x))' and so on...

[Dx]

Originally posted by The_Brain
How do you solve the derivative of x^x? I'm sure it's fairly easy-- I'm just beginning calc though and none of the "forumlas" work.
check out the power rule, learn it , know it, LOVE IT! Its easy, i am just playing but check it out, ok. Any problems with it lemme know, k.
Dx [;)]

[Tom Mattson]

Originally posted by Dx
[B]check out the power rule, learn it , know it, LOVE IT! Its easy, i am just playing but check it out, ok. Any problems with it lemme know, k.


No, the power rule is not applicable here because the exponent is not a constant. Stephen Privitera's solution is correct; go with that.