How can the derivative of x^x be solved?

[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=x^x
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=x^a, where a is a constant.
lny=alnx
dy/dx=y(a/x)=x^a(a/x)=ax^a-1
a formula I'm sure you know well

 

[Guybrush Threepwood]

or x^x = e^x*ln(x) ad continue from here...
(e^x*ln(x))' = e^x*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 let me know, k.
Dx

 

[quantumdude]

Originally posted by Dx
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 let me 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.