Computing derivative $x^{x^x}$ Could you show me how to compute $f'(x)$, where $f(x)=x^{x^x}$.
I know that for $g(x)=x^x=e^{x\ln x} \ \ $ $g'(x)=e^{x\ln x}(\ln x+1)$
Now, my problem is this: is $f(x)=x^{x^x}= e^{x^x \ln x}$ or $e^{x\ln x^x}$ ?

Since you know the derivative of $g(x)=x^x$, you can use the chain rule.
Then we can write $x^{x^x}=x^{g(x)}= e^{\ln (x^{g(x)})}=e^{g(x) \ln x}$.
Now use the chain rule (twice), and compute.
