# Find the values

Question:

Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, when

$\mathrm{x}=\frac{3 \text { at }}{1+\mathrm{t}^{2}}$ and $\mathrm{y}=\frac{3 \mathrm{at}^{2}}{1+\mathrm{t}^{2}}$

Solution:

$\mathrm{aS} \mathrm{X}=\frac{3 \mathrm{at}}{1+\mathrm{t}^{2}}$

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

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

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

$=\left[\frac{(3 a)+3 a t^{2}-6 a t^{2}}{\left(1+t^{2}\right)^{2}}\right]$

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

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

And $y=\frac{3 a t^{2}}{1+t^{2}}$

Differentiating it with respect to $t$ using quotion rule

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

$\frac{d y}{d t}=\left[\frac{\left(1+t^{2}\right)(6 a t)-\left(3 a t^{2}\right)(2 t)}{\left(1+t^{2}\right)^{2}}\right]$

$=\left[\frac{6 a t+6 a t^{3}-6 a t^{3}}{\left(1+t^{2}\right)^{2}}\right]$

$\frac{d y}{d t}=\frac{6 a t}{\left(1+t^{2}\right)^{2}} \cdots(2)$

Dividing equation (2) by (1),

$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{6 a t}{\left(1+t^{2}\right)^{2}} \times \frac{3 a\left(1-t^{2}\right)}{\left(1+t^{2}\right)^{2}}$

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