# What is the derivative of f(t) = (1-te^t, t^2+t ) ?

Jun 24, 2017

$f ' \left(t\right) = - \frac{2 t + 1}{\left(t + 1\right) {e}^{t}}$

#### Explanation:

We have:

$f \left(t\right) = \left(1 - t {e}^{t} , {t}^{2} + t\right)$

Which we can write as

$f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$

Where:

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

Then differentiating wrt $t$ (and applying the product rule) we have:

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

So then by the chain rule we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy} / \mathrm{dt}}{\mathrm{dx} / \mathrm{dt}}$

$\text{ } = \frac{2 t + 1}{- t {e}^{t} - {e}^{t}}$

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