# What is the derivative of f(t) = (t^3-e^(3t-1) , 3t^2+2e^t ) ?

Dec 30, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 t + 2 {e}^{t}}{3 {t}^{2} - 3 {e}^{3 t - 1}}$

#### Explanation:

$x ' \left(t\right) = 3 {t}^{2} - 3 {e}^{3 t - 1}$
$y ' \left(t\right) = 6 t + 2 {e}^{t}$

The derivative of the parametric function is

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y ' \left(t\right)}{x ' \left(t\right)} = \frac{6 t + 2 {e}^{t}}{3 {t}^{2} - 3 {e}^{3 t - 1}}$