differentiate dy/dx = x^x^x
Let u = (x^x)^x.
Let v = x^x.
Then u = v^x,
Also, ln(v) = x.ln(x).
Differentiating both sides wrt x,
(1/v).dv/dx = ln(x) + x/x
dv/dx = (x^x)(1 + ln(x))
From u = v^x,
ln(u) = x.ln(v)
Differentiating both sides wrt x,
(1/u).du/dx = ln(v) + x.(1/v).dv/dx
du/dx = u{ln(v) + (x/v).(x^x).(1 + ln(x))
du/dx = ((x^x)^x){ln(x^x) + x(1 + ln(x))}