Oh, it's perfectly possible, and fun too .
I have class soon, but I will get you started.
First, what is the derivative of x^x?
We have y = x^x
Note: x > 0
ln(y) = ln(x^x)
ln(y) = x*(ln(x))
Differentiate both sides with respect to x;
y'(1/y) = ln(x) + x*(1/x) = (ln(x) + 1)
y' = (ln(x) + 1)*y
But recall y = x^x
y' = (ln(x) + 1)*x^x
Now you can use this to find the derivative of x^x^x.