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

Mar 9, 2018

$\frac{\mathrm{dx}}{\mathrm{dt}} = - \frac{1}{t} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{2 t}{1 - {t}^{2}} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 {t}^{3}}{1 - {t}^{2}} ^ 2$

#### Explanation:

We have:

$x \left(t\right) = \frac{1}{t} \setminus \setminus$ and $\setminus \setminus y \left(t\right) = \frac{1}{1 - {t}^{2}}$

Differentiating each parametric term wrt $t$ we get:

$\frac{\mathrm{dx}}{\mathrm{dt}} = - \frac{1}{t} ^ 2$

And:

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

This is is technically the answer to the question, however more likely we would be asked to find $\frac{\mathrm{dy}}{\mathrm{dx}}$, which we obtain using the chain rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

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

$\setminus \setminus \setminus \setminus \setminus = \frac{- 2 {t}^{3}}{1 - {t}^{2}} ^ 2$