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

Feb 3, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{t} \left(t + 1\right) - 1}{2 {e}^{t} \left(t + 1\right) + 4 t}$

#### Explanation:

For a parametric function $f \left(x \left(t\right) , y \left(t\right)\right)$, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

As $x \left(t\right) = - t {e}^{t} + t$,

$\frac{\mathrm{dx}}{\mathrm{dt}} = - \left(t {e}^{t} + 1 \cdot {e}^{t}\right) + 1 = 1 - {e}^{t} \left(t + 1\right)$

and as $y \left(t\right) = - 2 {t}^{2} - 2 t {e}^{t}$, $\frac{\mathrm{dy}}{\mathrm{dt}} = - 4 t - 2 \left(t {e}^{t} + 1 \cdot {e}^{t}\right)$

= $- 4 t - 2 {e}^{t} \left(t + 1\right)$

Hence, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - {e}^{t} \left(t + 1\right)}{- 4 t - 2 {e}^{t} \left(t + 1\right)}$

= $\frac{{e}^{t} \left(t + 1\right) - 1}{2 {e}^{t} \left(t + 1\right) + 4 t}$