Question

Derivada de
$y=t-t^2cost$

Answer 1

$\\ y = t -t^2cost \\ \\ y' = 1+(-t^2)'cost +(-t^2)cost' \\ \\ y' = 1 -2tcost-t^2*(-sent) \\ \\ y' = 1-2tcost+t^2sent$

Answer 2

$\sf \displaystyle \:y=t-t^2cost\\\\\\\frac{d}{dt}\left(t-t^2\cos \left(t\right)\right)\\\\\\=\frac{d}{dt}\left(t\right)-\frac{d}{dt}\left(t^2\cos \left(t\right)\right)\\\\\\=1-\left(2t\cos \left(t\right)-t^2\sin \left(t\right)\right)\\\\\\\to \boxed{\sf =1-2t\cos \left(t\right)+t^2\sin \left(t\right)}$