Calculus  derivatives
posted by Lucy X on .
Okay, I want to find the derivative of (x^x)^(x^x)...
Well, I already went through the process of finding the derivative of just x^x (I used ln to do this).
This is what I found...
dy/dx = x^x(lnx+1)
So now I want to find the derivative of (x^x)^(x^x) (using ln).
This is what I did...
y = (x^x)^(x^x)
lny = x^xlnx^x
(1/y)(dy/dx) = (d/dx)(x^x)ln(x^x) + (d/dx)ln(x)^x*(x^x)
Yeah, I don't know if it is right.
Also, how do I find the derivative of ln(x)^x?
It seems simple enough, but I don't know...

let u=x^{x}
ln(u) = x ln(x)
(1/u)du/dx = ln(x) + 1
du/dx = u (ln(x) + 1)
du/dx = x^{x}(ln(x) + 1)
So this confirms what you've got for
d(x^{x})/dx
To find (x^x)^(x^x), it is not as complicated as it looks once we've got the derivative of x^{x}.
I would start with
u=x^{x} and
du/dx = x^{x}(ln(x) + 1)
Given
y=(x^x)^(x^x)
substitute
u=x^{x}, then
y=u^{u}
ln(y)=u ln(u)
dy/du=u^{u}(ln(u)+1)
dy/dx
=dy/du.du/dx
=u^{u}(ln(u)+1) . x^{x}(ln(x) + 1)
The rest is algebra to eliminate u from the right hand side to get
x^x^x^x(x^{x+1} ln(x)(ln(x)+1)+x^{x}(ln(x)+1))
or in any other equivalent form that you wish.