What is the first derivative according to x of the function: f(x)=x^(x)?

Question

parthkohli · Answer

The first derivative of any function $\large x^x$ is $\large x \times x^{x - 1}$

parthkohli · Answer

Well that just makes $\Large x^x$

anonymous · Answer

is it $$x^{2+x-1}$$??

parthkohli · Answer

Nope, it's just $\large x^x$

anonymous · Answer

Okay I think for y' = x^x you first need to take the ln of both sides to get 
lny = lnx^x 
then because of log properties you can rewrite the right side as 
lny=xlnx 
Now I think we do implicit differentiation 
(1/y)y' = x(1/x)+ln(x)
y' = y(1+ln(x))
and we know that y=x^x so
y'=x^x(1+ln(x)) or y'= x^x+x^xln(x) and I just checked wolfram and it's right so unless I am misunderstanding the question I am pretty sure Parthkohli is mistaken.