If yx=xx,x>0, then y"2-2y'2 is equal to :

Question

y (x) = x^{x}, x > 0

, then

y^{"} (2) - 2 y^{'} (2)

is equal to :

8 \log_{e} 2 - 2

4 \log_{e} 2 + 2

4 {(\log_{e} 2)}^{2} - 2

4 {(\log_{e} 2)}^{2} + 2

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