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

Jan 10, 2018

$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{{e}^{t} 4 {t}^{2} - 8 t {e}^{t}}{16 {t}^{4}}$

$\frac{\mathrm{dy}}{\mathrm{dt}} = 1 - {e}^{t}$

Explanation:

Both the equations would need to be differentiated seperatly as follows:

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

Therefore $\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{{e}^{t} 4 {t}^{2} - 8 t {e}^{t}}{16 {t}^{4}}$

$y \left(t\right) = t - {e}^{t}$

therefore $\frac{\mathrm{dy}}{\mathrm{dt}} = 1 - {e}^{t}$