Hi Calebius,

Suppose that y = x^g(x) and you want to differentiate y with respect to x. First take the natural log of both sides and use the properties of the log fuctions to get

ln(y) = ln( x^g(x)) = g(x) ln(x)

Now differentiate both sides

 ^y'/_y = g'(x) ln(x) + ^g(x)/_x

and hence

y' = y [ g'(x) ln(x) + ^g(x)/_x ] that is
y' = x^g(x) [ g'(x) ln(x) + ^g(x)/_x ]

In your problem g(x) = x^x so you are going to have to use this method again to find g'(x).

Penny