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

Jan 17, 2018

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

#### Explanation:

Let $x \left(t\right) = {e}^{t} / t + {e}^{t}$ and $y \left(t\right) = {e}^{t} - t \cos t$ Therefore, $f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$, so $f ' \left(t\right) = \left(x ' \left(t\right) , y ' \left(t\right)\right)$.

$x ' \left(t\right) = \frac{d}{\mathrm{dt}} \left[{e}^{t} / t + {e}^{t}\right]$
$x ' \left(t\right) = \frac{d}{\mathrm{dt}} \left[{e}^{t} / t\right] + \frac{d}{\mathrm{dt}} \left[{e}^{t}\right]$
$x ' \left(t\right) = \frac{t \frac{d}{\mathrm{dt}} \left[{e}^{t}\right] - {e}^{t} \frac{d}{\mathrm{dt}} \left[t\right]}{t} ^ 2 + {e}^{t}$
$x ' \left(t\right) = \frac{t {e}^{t} - {e}^{t}}{t} ^ 2 + {e}^{t}$
$x ' \left(t\right) = {e}^{t} \left(\frac{t - 1}{t} ^ 2 + 1\right)$

$y ' \left(t\right) = \frac{d}{\mathrm{dt}} \left[{e}^{t} - t \cos t\right]$
$y ' \left(t\right) = \frac{d}{\mathrm{dt}} \left[{e}^{t}\right] - \frac{d}{\mathrm{dt}} \left[t \cos t\right]$
$y ' \left(t\right) = {e}^{t} - \left(t \frac{d}{\mathrm{dt}} \left[\cos t\right] + \cos t \frac{d}{\mathrm{dt}} \left[t\right]\right)$
$y ' \left(t\right) = {e}^{t} - \left(- t \sin t + \cos t\right)$
$y ' \left(t\right) = {e}^{t} + t \sin t - \cos t$

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