# How do you differentiate the following parametric equation:  x(t)=-2te^t+4t, y(t)= -2t^2-3e^(t) ?

Jul 12, 2017

the first derivative, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} / \frac{\mathrm{dx}}{\mathrm{dt}}$ for parametric equations
the reason this holds true could be seen by treating $\mathrm{dy} , \mathrm{dx} , \mathmr{and} \mathrm{dt}$ as differentials (which they are) and upon dividing, $\mathrm{dt}$ cancels out and you are left with $\frac{\mathrm{dy}}{\mathrm{dx}}$

as a refresher, $\frac{d}{\mathrm{dt}} \left({x}^{t}\right) = t {x}^{t - 1}$ and $\frac{d}{\mathrm{dt}} \left({e}^{t}\right) = {e}^{t} \cdot t '$

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$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left(- 2 {t}^{2} - 3 {e}^{t}\right) = - 4 t - 3 {e}^{t}$

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

$= - 2 {e}^{t} - 2 t {e}^{t} + 4$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} / \frac{\mathrm{dx}}{\mathrm{dt}} = \frac{- 4 t - 3 {e}^{t}}{- 2 {e}^{t} - 2 t {e}^{t} + 4}$