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

Jan 5, 2016

Explanation is given below.

#### Explanation:

$x \left(t\right) = {e}^{{t}^{2} - t}$
$y \left(t\right) = {t}^{2} - t$

To find the derivative $\frac{\mathrm{dy}}{\mathrm{dx}}$ we need to find $y ' \left(t\right)$ and $x ' \left(t\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y ' \left(t\right)}{x ' \left(t\right)}$

$x \left(t\right) = {e}^{{t}^{2} - t}$
Differentiating with respect to $t$

$x ' \left(t\right) = {e}^{{t}^{2} - t} \frac{d}{\mathrm{dt}} \left({t}^{2} - t\right)$
$x ' \left(t\right) = {e}^{{t}^{2} - t} \left(2 t - 1\right)$
$x ' \left(t\right) = \left(2 t - 1\right) {e}^{{t}^{2} - t}$

$y \left(t\right) = {t}^{2} - t$
Differentiating with respect to $t$

$y ' \left(t\right) = 2 t - 1$

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

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