Find $\frac{d y}{d x}$, when
$x=\frac{2 t}{1+t^{2}}$ and $y=\frac{1-t^{2}}{1+t^{2}}$
We have, $x=\frac{2 t}{1+t^{2}}$
$\Rightarrow \frac{d x}{d t}=\left[\frac{\left(1+t^{2}\right) \frac{d}{d t}(2 t)-2 t \frac{d}{d t}\left(1+t^{2}\right)}{\left(1+t^{2}\right)^{2}}\right]$ [using quotient rule]
$\Rightarrow \frac{d x}{d t}=\left[\frac{\left(1+t^{2}\right)(2)-2 t(2 t)}{\left(1+t^{2}\right)^{2}}\right]$
$\Rightarrow \frac{d x}{d t}=\left[\frac{2+2 t^{2}-4 t^{2}}{\left(1+t^{2}\right)^{2}}\right]$
$\Rightarrow \frac{d x}{d t}=\left[\frac{2-2 t^{2}}{\left(1+t^{2}\right)^{2}}\right]$ .....(1)
and,
$y=\frac{1-t^{2}}{1+t^{2}}$
$\Rightarrow \frac{d y}{d t}=\left[\frac{\left(1+t^{2}\right) \frac{d}{d t}\left(1-t^{2}\right)-\left(1-t^{2}\right) \frac{d}{d t}\left(1+t^{2}\right)}{\left(1+t^{2}\right)^{2}}\right]$
$\Rightarrow \frac{d y}{d t}=\left[\frac{\left(1+t^{2}\right)(-2 t)-\left(1-t^{2}\right)(2 t)}{\left(1+t^{2}\right)^{2}}\right]$
$\Rightarrow \frac{d y}{d t}=\left[\frac{-4 t}{\left(1+t^{2}\right)^{2}}\right]$ ......(2)
Dividing equation (ii) by (i), we get,
$\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{-4 t}{\left(1+t^{2}\right)^{2}} \times \frac{\left(1+t^{2}\right)^{2}}{2\left(1-t^{2}\right)}$
$\Rightarrow \frac{d y}{d x}=\frac{-2 t}{1-t^{2}}$
$\Rightarrow \frac{d y}{d x}=-\frac{x}{y}$ $\left[\because \frac{x}{y}=\frac{2 t}{1+t^{2}} \times \frac{1+t^{2}}{1-t^{2}}=\frac{2 t}{1-t^{2}}\right]$
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