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

Nov 9, 2016

$\frac{t {e}^{t} + {e}^{t} - 2 t + 1}{1 + {e}^{t}}$

Explanation:

Differentiation states $\frac{\mathrm{dy}}{\mathrm{dx}}$

$x = t + {e}^{t}$
$y = t {e}^{t} - {t}^{2} + t$

Take the derivative of both functions
$x ' = 1 + {e}^{t}$
$y ' = \frac{d}{\mathrm{dt}} \left(t {e}^{t}\right) - \frac{d}{\mathrm{dt}} \left({t}^{2}\right) + \frac{d}{\mathrm{dt}} \left(t\right)$
Use product rule for $t {e}^{t}$, which states: f' g + f g'
$\frac{d}{\mathrm{dt}} \left(t\right) \cdot {e}^{t} + t \cdot \frac{d}{\mathrm{dt}} \left({e}^{t}\right) - 2 t + 1$

$1 \cdot {e}^{t} + t \cdot {e}^{t} - 2 t + 1$

$y ' = t {e}^{t} + {e}^{t} - 2 t + 1$

so now $\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{t {e}^{t} + {e}^{t} - 2 t + 1}{1 + {e}^{t}}$