# How do you differentiate the following parametric equation:  x(t)=t/(t-4), y(t)=1/(1-t^2) ?

Feb 28, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{t {\left(t - 4\right)}^{2}}{2 {\left(1 - {t}^{2}\right)}^{2}} = - \frac{t}{2} {\left(\frac{t - 4}{1 - {t}^{2}}\right)}^{2}$

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y ' \left(t\right)}{x ' \left(t\right)}$

$y \left(t\right) = \frac{1}{1 - {t}^{2}}$
$y ' \left(t\right) = \frac{\left(1 - {t}^{2}\right) \frac{d}{\mathrm{dt}} \left[1\right] - 1 \frac{d}{\mathrm{dt}} \left[1 - {t}^{2}\right]}{1 - {t}^{2}} ^ 2$
$\textcolor{w h i t e}{y ' \left(t\right)} = \frac{- \left(- 2 t\right)}{1 - {t}^{2}} ^ 2$
$\textcolor{w h i t e}{y ' \left(t\right)} = \frac{2 t}{1 - {t}^{2}} ^ 2$

$x \left(t\right) = \frac{t}{t - 4}$
$x ' \left(t\right) = \frac{\left(t - 4\right) \frac{d}{\mathrm{dt}} \left[t\right] - t \frac{d}{\mathrm{dt}} \left[t - 4\right]}{t - 4} ^ 2$
$\textcolor{w h i t e}{x ' \left(t\right)} = \frac{t - 4 - t}{t - 4} ^ 2$
$\textcolor{w h i t e}{x ' \left(t\right)} = - \frac{4}{t - 4} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 t}{1 - {t}^{2}} ^ 2 \div - \frac{4}{t - 4} ^ 2 = \frac{2 t}{1 - {t}^{2}} ^ 2 \times - {\left(t - 4\right)}^{2} / 4 = \frac{- 2 t {\left(t - 4\right)}^{2}}{4 {\left(1 - {t}^{2}\right)}^{2}} = - \frac{t {\left(t - 4\right)}^{2}}{2 {\left(1 - {t}^{2}\right)}^{2}} = - \frac{t}{2} {\left(\frac{t - 4}{1 - {t}^{2}}\right)}^{2}$