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If $x=\frac{1-t}{1+t}, y=\frac{2 t}{1+t}$, then $\frac{d^{2} y}{d x^{2}}=$
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We have, $x=\frac{1-t}{1+t}$
$$
\frac{d x}{d t}=\frac{-(1+t)-(1-t)}{(1+t)^{2}}=\frac{-2}{1+t^{2}} \text { and } y=\frac{2 t}{1+t}
$$
$$
\frac{d y}{d t}=\frac{(1+t) 2-2 t}{(1+t)^{2}}=\frac{2}{(1+t)^{2}}
$$
Now, $\quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=-1$ and $\quad \frac{d^{2} y}{d x^{2}}=0$
$$
\frac{d x}{d t}=\frac{-(1+t)-(1-t)}{(1+t)^{2}}=\frac{-2}{1+t^{2}} \text { and } y=\frac{2 t}{1+t}
$$
$$
\frac{d y}{d t}=\frac{(1+t) 2-2 t}{(1+t)^{2}}=\frac{2}{(1+t)^{2}}
$$
Now, $\quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=-1$ and $\quad \frac{d^{2} y}{d x^{2}}=0$
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