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

Dec 31, 2015

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

#### Explanation:

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

The derivative of the parametric function is

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y ' \left(t\right)}{x ' \left(t\right)} = \frac{{e}^{t} - 2 t + 1}{1 + {e}^{t} + t {e}^{t}}$