# How do you differentiate the following parametric equation:  x(t)=te^t+5t, y(t)= t^2-3e^(t) ?

Sep 23, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(t + 1\right) {e}^{t} + 5}{2 t - 3 {e}^{t}} .$

#### Explanation:

By Defn., $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

$y = y \left(t\right) = {t}^{2} - 3 {e}^{t} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left({t}^{2} - 3 {e}^{t}\right) = \frac{d}{\mathrm{dt}} \left({t}^{2}\right) - \frac{d}{\mathrm{dt}} \left(3 {e}^{t}\right)$

$= 2 t - 3 {e}^{t}$

$x = x \left(t\right) = t {e}^{t} + 5 t \Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left(t {e}^{t} + 5 t\right) = \frac{d}{\mathrm{dt}} \left(t {e}^{t}\right) + \frac{d}{\mathrm{dt}} \left(5 t\right)$

$= t \frac{d}{\mathrm{dt}} \left({e}^{t}\right) + {e}^{t} \frac{d}{\mathrm{dt}} \left(t\right) + 5 = t {e}^{t} + {e}^{t} + 5 = \left(t + 1\right) {e}^{t} + 5$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(t + 1\right) {e}^{t} + 5}{2 t - 3 {e}^{t}} .$