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December 21, 2014

December 21, 2014

Posted by **Lucy X** on Sunday, November 29, 2009 at 3:58am.

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 -
**MathMate**, Sunday, November 29, 2009 at 9:03amlet 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.

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