If $x = \frac{a}{1+t}$ and $y =\frac{a}{(1+t)^2}$ where $a > 0$, then $\frac{d^2y}{dx^2}$ at $t = 1$ is

$\frac{2}{a}$

The correct answer is Option (2) → $\frac{2}{a}$ **

$x=\frac{a}{1+t}$

$y=\frac{a}{(1+t)^2}$

$\frac{dx}{dt}=-a(1+t)^{-2}$

$\frac{dy}{dt}=-2a(1+t)^{-3}$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-2a(1+t)^{-3}}{-a(1+t)^{-2}}=\frac{2}{1+t}$

$\frac{d}{dt}\left(\frac{dy}{dx}\right)=-2(1+t)^{-2}$

$\frac{dt}{dx}=\frac{1}{\frac{dx}{dt}}=-\frac{(1+t)^2}{a}$

$\frac{d^2y}{dx^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot\frac{dt}{dx} =\left[-2(1+t)^{-2}\right]\left[-\frac{(1+t)^2}{a}\right]=\frac{2}{a}$

$\frac{d^2y}{dx^2}\bigg|_{t=1}=\frac{2}{a}$