# Solve this

Question:

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

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

Solution:

We have, $y=\frac{2 t}{1+t^{2}}$

$\Rightarrow \frac{d y}{d t}=\left[\frac{\left(1+t^{2}\right) \frac{d}{d t}(2 t)-2 t \frac{d}{d t}\left(1+t^{2}\right)}{\left(1+t^{2}\right)^{2}}\right]$        [using quotient rule]

$\Rightarrow \frac{d y}{d t}=\left[\frac{\left(1+t^{2}\right)(2)-2 t(2 t)}{\left(1+t^{2}\right)^{2}}\right]$

$\Rightarrow \frac{d y}{d t}=\left[\frac{2+2 t^{2}-4 t^{2}}{\left(1+t^{2}\right)^{2}}\right]$

$\Rightarrow \frac{d y}{d t}=\left[\frac{2-2 t^{2}}{\left(1+t^{2}\right)^{2}}\right]$               ......(1)

and,

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

$\Rightarrow \frac{d x}{d t}=\left[\frac{\left(1+t^{2}\right) \frac{d}{d t}\left(1-t^{2}\right)-\left(1-t^{2}\right) \frac{d}{d t}\left(1+t^{2}\right)}{\left(1+t^{2}\right)^{2}}\right]$

$\Rightarrow \frac{d x}{d t}=\left[\frac{\left(1+t^{2}\right)(-2 t)-\left(1-t^{2}\right)(2 t)}{\left(1+t^{2}\right)^{2}}\right]$

$\Rightarrow \frac{d x}{d t}=\left[\frac{-4 t}{\left(1+t^{2}\right)^{2}}\right]$              ....(2)

Dividing equation (i) by (ii), we get,

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

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

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