# Solve this

Question:

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

If $x=a\left(\frac{1+t^{2}}{1-t^{2}}\right)$ and $y=\frac{2 t}{1-t^{2}}$, find $\frac{d y}{d x}$

Solution:

Here,

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

differentiating bove function with respect to $t$, we have,

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

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

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

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

And

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

differentiating bove function with respect to $t$, we have,

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

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

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

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

$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{2\left[\frac{1+t^{2}}{1-t^{2}}\right]}{\frac{4 a t}{\left(1-t^{2}\right)^{2}}} \mid$ from equation 1 and 2

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1-\mathrm{t}^{4}}{2 \mathrm{at}}$