# What is the derivative of f(t) = (t^2-2e^(3t-1) , t-e^t ) ?

$f ' \left(t\right) = 2 t - 2 \left(3\right) {e}^{3 t - 1} , 1 - {e}^{t}$
$f \left(x\right) = \left({t}^{2} - 2 {e}^{3 t - 1} , t - {e}^{t}\right)$
first use the power rule for ${t}^{2} , t$
$f ' \left(x\right) = 2 {t}^{2 - 1} - 2 \frac{\mathrm{dy}}{\mathrm{dt}} \left(3 t - 1\right) {e}^{3 t - 1} , \frac{\mathrm{dy}}{\mathrm{dt}} \left(t\right) {e}^{t}$
then use the $\frac{\mathrm{dy}}{\mathrm{dx}} {\left(e\right)}^{u} = {e}^{u} \left(\mathrm{du}\right)$
therefore the derivative is $f ' \left(x\right) = 2 t - 6 {e}^{3 t - 1} , 1 - {e}^{t}$