If $x = \frac{1-t}{1+t}$ and $y = \frac{3t}{1+t}$, then $\frac{d^2y}{dx^2}$ is equal to

Given: $x = \frac{1-t}{1+t}$ and $y = \frac{3t}{1+t}$

Both $x$ and $y$ are given in terms of $t$, so:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

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For $x = \frac{1-t}{1+t}$:

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

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

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

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For $y = \frac{3t}{1+t}$:

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

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

$= \frac{3}{(1+t)^2}$

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$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

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

$= \frac{3}{-2}$

$= -\frac{3}{2}$

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Since $\frac{dy}{dx} = -\frac{3}{2}$ is a constant:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{3}{2}\right) = 0$

Therefore, $\frac{d^2y}{dx^2} = 0$