If $x=\frac{1}{t^2}, y =\frac{1}{t}$, then value of $\frac{d^2y}{dx^2}$ at $t=2$  is :

-2

The correct answer is Option (1) → -2

$x=\frac{1}{t^2}, y =\frac{1}{t}$

$⇒\frac{dx}{dt}=\frac{-2}{t^3},⇒\frac{dy}{dt}=\frac{-1}{t^2}$

$\frac{dy}{dx}=\frac{-1}{t^2}×\frac{t^3}{-2}=\frac{t}{2}$

$⇒\frac{d^2y}{dx^2}=\frac{1}{2}×\frac{t^3}{-2}=\frac{t^3}{-4}$

$⇒\left.\frac{d^2y}{dx^2}\right|_{t=2}=\frac{8}{-4}=-2$