If y(x) = x ^ x , x > 0 , then y ^ " (2) - 2 y ^ ' (2) is equal to :

Given, $y = {x}^{x}$

$⇒ \ln \left(y\right) = xln$\left(x\right)

Differentiate both sides w.r.t. $x$

\dfrac{1}{y} . y ' = 1 + \ln \left(x\right)

$⇒ y ' = y \left(1 + \ln x\right)$ . . . . . . . ( 1 )$</p><p>$⇒ y ' = {x}^{x}\left(1 + \ln x\right)$</p><p>at$x = 2 ⇒ y = 4$</p><p>So,$y ' ( 2 ) = 4\left(1 + \ln 2\right) . . . . . . . ( 2 )$</p><p>Now differentiate the equation \left(1\right)$,

$y " = y ' \left(1 + \ln x\right) + \dfrac{y}{x}$

$⇒ y " ( 2 ) = y ' \left(1 + \ln 2\right) + 2$

$⇒ y " ( 2 ) - y ' ( 2 ) = y ' ( \ln 2 ) + 2$

$⇒ y " ( 2 ) - 2y ' ( 2 ) = \left(\ln 2 - 1\right)y ' ( 2 ) + 2$

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

$= 4{\left(\ln 2\right)}^{2} - 2$