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