If $y=\frac{1}{x+1}$, then $\frac{d^2 y}{d x^2}$ at x = 2 is:

$\frac{2}{9}$ $\frac{3}{2}$ $\frac{2}{27}$ $\frac{3}{8}$

$\frac{2}{27}$

$y=\frac{1}{x+1}$ so differentiating y wrt x $\frac{dy}{d x} = \frac{-1}{(x+1)^2}$ again differentiating wrt x $\frac{d^2 y}{d x^2} = \frac{2}{(x+1)^3}$ so  $\left.\frac{d^2y}{dx^2}\right]_{x=2} = \frac{2}{(2+1)^3} = \frac{2}{3^3}$ = $\frac{2}{27}$