作业帮设y=e^(-x)cos(3-x),求dy 详细些
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设y=e^(-x)cos(3-x),求dy 详细些
设y=e^(-x)cos(3-x),求dy 详细些
设y=e^(-x)cos(3-x),求dy 详细些
∵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
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
书中原题啊
dy=[e^(-x)][sin(3-x)-cos(3-x)]dx
dy=[-e^(-x)cos(3-x)+sin(3-x)e^(-x)]dx
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