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

Mar 22, 2017

 dy/dx = (2t^2(1 - 3e^(3t))e^(-t)) /((t-1)

#### Explanation:

I assume that you want more than just the derivatives of the parametric equation, and actually require $\frac{\mathrm{dy}}{\mathrm{dx}}$

We have;

$x \left(t\right) = {e}^{t} / \left(t + t\right) = {e}^{t} / \left(2 t\right)$

Differentiating wrt $t$, and applying the quotient rule, we have;

$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{\left(2 t\right) \left({e}^{t}\right) - \left(2\right) \left({e}^{t}\right)}{2 t} ^ 2$
$\text{ } = \frac{2 \left(t - 1\right) {e}^{t}}{4 {t}^{2}}$
$\text{ } = \frac{\left(t - 1\right) {e}^{t}}{2 {t}^{2}}$

And:

$y \left(t\right) = t - {e}^{3 t}$

Differentiating wrt $t$, and applying the chain rule, we have;

$\frac{\mathrm{dy}}{\mathrm{dt}} = 1 - 3 {e}^{3 t}$

And finally by the chain rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy} / \mathrm{dt}}{\mathrm{dx} / \mathrm{dt}}$
$\text{ } = \frac{1 - 3 {e}^{3 t}}{\frac{\left(t - 1\right) {e}^{t}}{2 {t}^{2}}}$
$\text{ } = \left(1 - 3 {e}^{3 t}\right) \cdot \frac{2 {t}^{2}}{\left(t - 1\right) {e}^{t}}$
 " " = (2t^2(1 - 3e^(3t))e^(-t)) /((t-1)