Answer
$f'(t)=e^{at}(a\sin bt+b\cos bt)$
Work Step by Step
$f(t)=e^{at}\sin bt$ (Here, $a$ and $b$ are constants)
Differentiate using the product rule:
$f'(t)=(e^{at})(\sin bt)'+(\sin bt)(e^{at})'=...$
Now, use the chain rule to find both $(\sin bt)'$ and $(e^{at})'$:
$...=(e^{at})[(\cos bt)(bt)']+(\sin bt)[(e^{at})(at)']=...$
$...=(e^{at})(b\cos bt)+(\sin bt)(ae^{at})=ae^{at}\sin bt+be^{at}\cos bt=...$
Take out common factor $e^{at}$ to present a better looking answer: (Optional)
$f'(t)=e^{at}(a\sin bt+b\cos bt)$
