# Calculus - derivatives

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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...

• Calculus - derivatives -

let u=xx
ln(u) = x ln(x)
(1/u)du/dx = ln(x) + 1
du/dx = u (ln(x) + 1)
du/dx = xx(ln(x) + 1)

So this confirms what you've got for
d(xx)/dx

To find (x^x)^(x^x), it is not as complicated as it looks once we've got the derivative of xx.

u=xx and
du/dx = xx(ln(x) + 1)

Given
y=(x^x)^(x^x)
substitute
u=xx, then
y=uu
ln(y)=u ln(u)
dy/du=uu(ln(u)+1)

dy/dx
=dy/du.du/dx
=uu(ln(u)+1) . xx(ln(x) + 1)
The rest is algebra to eliminate u from the right hand side to get
x^x^x^x(xx+1 ln(x)(ln(x)+1)+xx(ln(x)+1))
or in any other equivalent form that you wish.