# How do you differentiate the following parametric equation:  (t-5t^3,3t^4-t^3)?

Dec 31, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{12 {t}^{3} - 3 {t}^{2}}{1 - 15 {t}^{2}}$

#### Explanation:

$x ' \left(t\right) = 1 - 15 {t}^{2}$
$y ' \left(t\right) = 12 {t}^{3} - 3 {t}^{2}$

The derivative of the parametric function is

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y ' \left(t\right)}{x ' \left(t\right)} = \frac{12 {t}^{3} - 3 {t}^{2}}{1 - 15 {t}^{2}}$