## Calculus (3rd Edition)

$$y''=2\sin t+4t \cos t-t^2\sin t,$$ $$y'''=6\cos t-6t\sin t-t^2\cos t.$$
Since $y=t^2\sin t$, then we have $y'=2t \sin t+t^2\cos t=t(2\sin t+t\cos t)$ and by the product rule, we get $$y''= 2\sin t+t\cos t+t(3\cos t-t\sin t)=2\sin t+4t \cos t-t^2\sin t,$$ and again by using the product rule, we have $$y'''=2\cos t+4 \cos t-4t\sin t-2t\sin t-t^2\cos t=6\cos t-6t\sin t-t^2\cos t.$$