# How do you differentiate the following parametric equation:  x(t)=(t+1)e^t, y(t)= (t-2)^2+t?

Jul 26, 2017

$\frac{2 t - 3}{{e}^{t} t + 2 {e}^{t}}$

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} / \frac{\mathrm{dx}}{\mathrm{dt}}$

since $\frac{\mathrm{dy}}{\mathrm{dt}} / \frac{\mathrm{dx}}{\mathrm{dt}} = \frac{\mathrm{dy}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}}$ (cancelling $\mathrm{dt}$ in the numerator and denominator)

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

$y \left(t\right) = {\left(t - 2\right)}^{2} + t$
$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left({\left(t - 2\right)}^{2} + t\right) = 2 \left(t - 2\right) + 1$

Therefore,
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \left(t - 2\right) + 1}{\left(t + 1\right) {e}^{t} + {e}^{t}} = \frac{2 t - 3}{{e}^{t} t + 2 {e}^{t}}$