If $x=\frac{1-t}{1+t}$ and $y=\frac{2t}{1+t}$ then $\frac{d^2y}{dx^2}$ is equal to

0

The correct answer is Option (2) → 0

$x=\frac{1-t}{1+t},\quad y=\frac{2t}{1+t}$

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

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

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

$\frac{d^2y}{dx^2}=\frac{d}{dx}(-1)=0$

Final answer: $0$