# How do you differentiate the following parametric equation:  (t-t^3,3/t^4-t^3)?

May 24, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {t}^{7} + 12}{3 {t}^{7} - {t}^{5}}$

#### Explanation:

In such a parametric equation, where $\left(x , y\right)$ are given by $\left(t - {t}^{3} , \frac{3}{t} ^ 4 - {t}^{3}\right)$

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

Now $\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left(\frac{3}{t} ^ 4 - {t}^{3}\right) = \frac{d}{\mathrm{dt}} \left(3 {t}^{- 4} - {t}^{3}\right)$

= $- 12 {t}^{- 5} - 3 {t}^{2} = - \frac{12}{t} ^ 5 - 3 {t}^{2}$

and $\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left(t - {t}^{3}\right) = 1 - 3 {t}^{2}$

Hence, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- \frac{12}{t} ^ 5 - 3 {t}^{2}}{1 - 3 {t}^{2}} = \frac{- 12 - 3 {t}^{7}}{{t}^{5} - 3 {t}^{7}} = \frac{3 {t}^{7} + 12}{3 {t}^{7} - {t}^{5}}$