# What is the derivative of f(t) = (t^3-te^(1-t) , t^2-6t+e^t ) ?

Aug 12, 2018

$f ' \left(t\right) = \left(x ' \left(t\right) , y ' \left(t\right)\right) = \left(\left[3 {t}^{2} + t {e}^{1 - t} - {e}^{1 - t}\right] , 2 t - 6 + {e}^{t}\right)$

#### Explanation:

Recall that to compue the derivative of the function $f \left(t\right)$, we just need to compute the derivatives of each parametric equation.

If we have $f \left(x , y\right) = \left(x , y\right)$, then $f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$, where $x \left(t\right)$ and $y \left(t\right)$ are our two parametric equations. These equations describe $x$ and $y$ as functions of the single parameter $t$.

For our particular example:
$x \left(t\right) = {t}^{3} - t {e}^{1 - t}$
$y \left(t\right) = {t}^{2} - 6 t + {e}^{t}$

So, we just need to take the derivative of each equation:

$x ' \left(t\right) = 3 {t}^{2} - \left(- t {e}^{1 - t} + {e}^{1 - t}\right)$
$y ' \left(t\right) = 2 t - 6 + {e}^{t}$

Our final solution is:

$f ' \left(t\right) = \left(x ' \left(t\right) , y ' \left(t\right)\right) = \left(\left[3 {t}^{2} + t {e}^{1 - t} - {e}^{1 - t}\right] , 2 t - 6 + {e}^{t}\right)$

Hope that helped :)