Question
If $y(x) = {\left( {{x^x}} \right)^x},\,x > 0$, then ${{{d^2}x} \over {d{y^2}}} + 20$ at x = 1 is equal to ____________.
Correct Answer
16
If $y(x) = {\left( {{x^x}} \right)^x},\,x > 0$, then ${{{d^2}x} \over {d{y^2}}} + 20$ at x = 1 is equal to ____________.