Find f(x)' of f(x)=x^x^x
I logged both sides to get
ln y = x^x ln x
But I am not sure what step to do after that.
You are assuming y = f(x)
lny = x*(x^x)
(1/y)*dy/dx = x^x + x*(x^x)*(1 + lnx)
f'(x) = dy/dx
= f(x)*[x^x + x*(x^x)*(1 + lnx)]
= x^x^x*[x^x + x*(x^x)*(1 + lnx)]
I had to look up the derivative of x^x.