Question

已知y=e^-xsin2x，求微分dy．

Answer 1

dy=（e^-xsin2x）'dx
=[e^-x（sin2x）'+（e^-x）'sin2x]dx
=（2e^-xcos2x-e^-xsin2x）dx
=e^-x（2cos2x-sin2x）dx．