If $\mathrm{y}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{x}},\mathrm{x}>0,\text{ then }{\mathrm{y}}^{\prime \prime }\left(2\right)-2{\mathrm{y}}^{\mathrm{\prime }}\left(2\right)$ is equal to :

$\mathrm{y}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{x}};\mathrm{x}>0$
$\begin{array}{l}{y}^{\text{'}}={\mathrm{x}}^{\mathrm{x}}(1+\log \mathrm{x})\\ {\mathrm{y}}^{\text{'}\text{'}}={\mathrm{x}}^{\mathrm{x}}(1+\log \mathrm{x})+{\mathrm{x}}^{\mathrm{x}}(1+\log \mathrm{x})\log \mathrm{x}+{\mathrm{x}}^{\mathrm{x}}\times \frac{1}{\mathrm{x}}\\ {\mathrm{y}}^{\text{'}\text{'}}={\mathrm{x}}^{\mathrm{x}}(1+\log \mathrm{x}{)}^2+{\mathrm{x}}^{\mathrm{x}-1}\\ \therefore {\mathrm{y}}^{\text{'}\text{'}}\left(2\right)-2y^{\text{'}}\left(2\right)=4(1+\log 2{)}^2+2-2(4(1+\log 2))\\ =4(\log 2{)}^2-2\end{array}$