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

Jan 3, 2018

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

#### Explanation:

Take the derivative component-by-component:

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

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

$= \left(2 t , 1 - {e}^{1 - t} \left(- 1\right)\right)$ (don't forget the chain rule!)

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