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设y=e^(-x)cos(3-x),求dy 详细些
人气:322 ℃ 时间:2020-06-20 12:02:40
解答
∵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
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