## Calculus: Early Transcendentals 8th Edition

$f'(t)=\pi t\cos\pi t+\sin\pi t$
$f(t)=t\sin\pi t$ Differentiate using the product rule: $f'(t)=(t)(\sin\pi t)'+(\sin\pi t)(t)'=...$ Now, use the chain rule to find $(\sin\pi t)'$: $...=(t)[(\cos\pi t)(\pi t)']+\sin\pi t=(t)(\pi\cos\pi t)+\sin\pi t=...$ $...=\pi t\cos\pi t+\sin\pi t$