# Solve this

Question:

Find $\frac{d y}{d x}$, when

$x=\frac{2 t}{1+t^{2}}$ and $y=\frac{1-t^{2}}{1+t^{2}}$

Solution:

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,

$\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{-4 t}{\left(1+t^{2}\right)^{2}} \times \frac{\left(1+t^{2}\right)^{2}}{2\left(1-t^{2}\right)}$

$\Rightarrow \frac{d y}{d x}=\frac{-2 t}{1-t^{2}}$

$\Rightarrow \frac{d y}{d x}=-\frac{x}{y}$              $\left[\because \frac{x}{y}=\frac{2 t}{1+t^{2}} \times \frac{1+t^{2}}{1-t^{2}}=\frac{2 t}{1-t^{2}}\right]$