Answer
$3\pi e^{-2x} \cos ( 3\pi x) -2 e^{-2x} \sin ( 3\pi x)$
Work Step by Step
Let us suppose that $f(x)= e^{-2x} \sin ( 3\pi x)$
Use $\dfrac {d}{dx} (e^a)=e^a \dfrac{da}{dx}$
We differentiate both sides with respect to $x$.
$f^{\prime}(x)=\dfrac{d}{dx} [ e^{-2x} \sin ( 3\pi x)] \\=e^{-2x} \cos ( 3\pi x) \times 3\pi+\sin (3\pi x) (e^{-2x} \times (-2)$
Simplify to obtain:
$f^{\prime}(x)=3\pi e^{-2x} \cos ( 3\pi x) -2 e^{-2x} \sin ( 3\pi x)$
