# Solve this

Question:

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

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

Solution:

$a s, x=\frac{2 t}{1+t^{2}}$

Differentiating it with respect to $t$ using quotient rule,

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

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

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

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

$\frac{\mathrm{dx}}{\mathrm{dt}}=\left[\frac{2-2 \mathrm{t}^{2}}{\left(1+\mathrm{t}^{2}\right)^{2}}\right]$      ......(1)

And, $\mathrm{y}=\frac{1-\mathrm{t}^{2}}{1+\mathrm{t}^{2}}$

Differentiating it with respect to $t$ using quotient rule,

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

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

$=\left[\frac{-2 t-2 t^{3}-2 t+2 t^{3}}{\left(1+t^{2}\right)^{2}}\right]$

$\frac{\mathrm{dy}}{\mathrm{dt}}=\left[\frac{-4 \mathrm{t}}{\left(1+\mathrm{t}^{2}\right)^{2}}\right] \cdots(2)$

dividing equation (2)by (1),

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

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

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