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

Aug 31, 2016

$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{{e}^{t}}{4 {t}^{2}} - \frac{{e}^{t}}{2 {t}^{3}} - 1$, $\frac{\mathrm{dy}}{\mathrm{dt}} = 1 - {e}^{t}$

#### Explanation:

Because the curve is expressed in terms of two functions of $t$ we can find the answer by differentiating each function individually with respect to $t$. First note that the equation for $x \left(t\right)$ can be simplified to:

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

While $y \left(t\right)$ can be left as:

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

Looking at $x \left(t\right)$, it is easy to see that the application of the product rule will yield a quick answer. While $y \left(t\right)$ is simply standard differentiation of each term. We also use the fact that $\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$.

$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{{e}^{t}}{4 {t}^{2}} - \frac{{e}^{t}}{2 {t}^{3}} - 1$

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