设y=e^(-x)cos(3-x),求dy 详细些

网友回答

∵dy/dx=y'=[e^(-x)]'cos(3-x)+e^(-x)[cos(3-x)]'
=-e^(-x)cos(3-x)+e^(-x)(sin(3-x)
=[-cos(3-x)+(sin(3-x)]e^(-x)
∴dy=e^(-x)[-cos(3-x)+(sin(3-x)]dx
======以下答案可供参考======
供参考答案1:
解:y=e^(-x)cos(3-x)
y'=-e^(-x)cos(3-x)+e^(-x)(-sin(3-x))(-1)
=e^(-x)(sin(3-x))-e^(-x)cos(3-x)
所以dy=[e^(-x)(sin(3-x))-e^(-x)cos(3-x)]dx
供参考答案2:
书中原题啊dy=[e^(-x)][sin(3-x)-cos(3-x)]dx
供参考答案3:
dy=[-e^(-x)cos(3-x)+sin(3-x)e^(-x)]dx