设y=e^(-x)cos(3-x),求dy 详细些

分类: 作业答案 • 2022-03-05 17:45:54

问题描述：

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