Answer
$f'(t)=e^{at}(bcosbt)+a sinbt(e^{at})$
Work Step by Step
$f'(t)=\frac{d}{dt}[e^{at}sinbt]$
$=e^{at}\frac{d}{dt}[sinbt]+sinbt\frac{d}{dx}[e^{at}]$
Hence, $f'(t)=e^{at}(bcosbt)+a sinbt(e^{at})$
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