If y = ( x)/(x) , then d^ 2 y dx^ 2 = 2 x^ 3 ( x - λ) , where λ is

Using quotient& rule, we get,
$\dfrac{dy}{dx} = \dfrac{1 - \ln x}{x^{2}}<br />$∴ \dfrac{d^{2}y}{dx^{2}} = \dfrac{\left(x^{2}\right)\left(- \dfrac{1}{x}\right) - \left(1 - \ln x\right)\left(2x\right)}{x4}$<br />$= \dfrac{- 1 - 2 + 2\ln x}{x^{3}}$<br />$= \dfrac{2}{x^{3}} \left(\ln x - \left(1 + $$\dfrac{1}{2}$$\right)\right)