# How do you differentiate the following parametric equation:  x(t)=-te^t+t, y(t)= 3t^2+2t ?

Jul 1, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 t + 2}{1 - {e}^{t} \left(1 + t\right)}$

#### Explanation:

When parametric equations are given $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

Now as $y \left(t\right) = 3 {t}^{2} + 2 t$ and $x \left(t\right) = - t {e}^{t} + t$

$\frac{\mathrm{dy}}{\mathrm{dt}} = 6 t + 2$ and

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

Hence

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 t + 2}{1 - {e}^{t} \left(1 + t\right)}$