If `y=x^x^x^(((dotoo)))` , find `(dy)/(dx)` .

Explanation

Given that
`y=x^x^x^(((dotoo)))`
`y=x^(y)`
`"or "log y = y log x`
`"or "(1)/(y)(dy)/(dx)=(dy)/(dx)xxlog x+y(1)/(x)" "[Diff. both sides w.r.t.x]`
`"or "(dy)/(dx)({1-y log x })/(y)=(y)/(x)`
`"or "(dy)/(dx)=(y^(2))/(x(1-y log x))`