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

May 7, 2017

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

Explanation:

Take the derivative of the $x$ component and $y$ component separately.

That is, if $f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$ then $f ' \left(t\right) = \left(x ' \left(t\right) , y ' \left(t\right)\right)$.

So:

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

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

And

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

$y ' \left(t\right) = {e}^{t} + \sin t$

So:

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