Derivative of $x^{x^x}$ I need to compute the derivative of $$z=x^{x^x}$$
I made it like this:
$$y=x^x$$ so $$z=y^x$$
then $$\ln(z)=x\ln(y)$$ 
taking derivative from both sides:
$$\frac{z'}{z}=\ln(y)+x\left(\frac{1}{y}\right)y'$$
I know that $$(x^x)'=x^x(1+\ln(x))$$
So : $$z'=z*(\ln(x^x)+x*\left(\frac{1}{x^x}\right)*(x^x)'))=x^{x^x}(x\ln(x)+x\ln(x)+x)=x^{x^x}\left(2x\ln(x)+x\right)$$ I know I did something wrong. Where is my Mistake??
Thanks alot!

I think your confusion stems from what you think $x^{x^x}$ means.
$x^{x^x}$ means taking $x$ to the power of $x^x$ which is distinctly different from taking $x^x$ to the power of $x$.
As an example, $3^{3^3} = 3^{(3^3)} = 3^{27}$. 
$3^{3^3} \neq (3^3)^3=27^3$.
So $z \neq y^x$. Instead, $z = x^y$.
