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

Dec 30, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 t - 1}{2 t {e}^{{t}^{2}}}$

#### Explanation:

$x ' \left(t\right) = 2 t {e}^{{t}^{2}}$
$y ' \left(t\right) = 4 t - 1$

The derivative of the parametric equation is

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y ' \left(t\right)}{x ' \left(t\right)} = \frac{4 t - 1}{2 t {e}^{{t}^{2}}}$