If y(x) = (x^x^x), x > 0 then d^2x/dy^2 + 20 at x = 1 is equal to

Question

If y(x) =(\(x^{x^x}\)), x>0 then \(\frac{d^2{x}}{dy^2}+20\, at \,x=1\,is\,equal\,to\, \colon\)

Answer 1

Correct answer: 16