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

Feb 1, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{t} + t \sin t}{1 + {e}^{t}}$

Explanation:

$x = t + {e}^{t}$
$y = {e}^{t} - t \cos t$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} \div \frac{\mathrm{dx}}{\mathrm{dt}}$

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left[{e}^{t} - t \cos t\right]$
$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dt}}} = \frac{d}{\mathrm{dt}} \left[{e}^{t}\right] - \frac{d}{\mathrm{dt}} \left[t \cos t\right]$
$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dt}}} = {e}^{t} + t \sin t$

$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left[t + {e}^{t}\right]$
$\textcolor{w h i t e}{\frac{\mathrm{dx}}{\mathrm{dt}}} = \frac{d}{\mathrm{dt}} \left[t\right] + \frac{d}{\mathrm{dt}} \left[{e}^{t}\right]$
$\textcolor{w h i t e}{\frac{\mathrm{dx}}{\mathrm{dt}}} = 1 + {e}^{t}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{t} + t \sin t}{1 + {e}^{t}}$