Given, y=xx
⇒ln(y)=xln(x)
Differentiate both sides w.r.t. x
y1.y′=1+ln(x)
⇒y′=y(1+lnx).......(1)
⇒y′=xx(1+lnx)
at x=2⇒y=4
So, y′(2)=4(1+ln2).......(2)
Now differentiate the equation (1),
y"=y′(1+lnx)+xy
⇒y"(2)=y′(1+ln2)+2
⇒y"(2)−y′(2)=y′(ln2)+2
⇒y"(2)−2y′(2)=(ln2−1)y′(2)+2
=4(ln2−1)(ln2+1)+2
=4(ln2)2−2