Given function

$y=x^x, \, x>0$

We need to find:

$y''(2)-2y'(2)$

Differentiate using logarithmic differentiation

Let $y=x^x$ Taking natural log: $\ln y = x\ln x$

Differentiate both sides w.r.t. $x$: $\frac{y'}{y} = \ln x + 1$

Hence, $y' = y(\ln x+1)=x^x(\ln x+1)$

Find second derivative

Differentiate $y'=x^x(\ln x+1)$

Using product rule: $y'' = (x^x)'(\ln x+1) + x^x\left(\frac{1}{x}\right)$

Since $(x^x)'=x^x(\ln x+1),$ we get $y'' = x^x(\ln x+1)^2 + x^x\frac{1}{x}$

So, $y'' = x^x\left[(\ln x+1)^2+\frac{1}{x}\right]$

Evaluate at $x=2$

First, $y'(2)=2^2(\ln 2+1)=4(\ln 2+1)$

Next, $y''(2)=2^2\left[(\ln 2+1)^2+\frac{1}{2}\right]$ $=4(\ln 2+1)^2+2$

Compute $y''(2)-2y'(2)$

$y''(2)-2y'(2)=\left[4(\ln 2+1)^2+2\right]-2\cdot 4(\ln 2+1)$

$=4(\ln 2+1)^2+2-8(\ln 2+1)$

Expand: $4\left((\ln 2)^2+2\ln 2+1\right)+2-8\ln 2-8$

$=4(\ln 2)^2+8\ln 2+4+2-8\ln 2-8$

$=4(\ln 2)^2-2$

Match with options

$4(\log_e 2)^2-2$

This is Option D.