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求xCosx的n阶导数,大神你在哪里啊

解:y=xcosx y`=cosx-xsinx y``=-2sinx-xcosx y```=-3cosx+xsinx y````=4sinx-xcosx 于是 y(n)=nsin(x+nπ/2)+xsin[x+(n+1)π/2]

解:y=xcosx y`=cosx-xsinx y``=-2sinx-xcosx y```=-3cosx+xsinx y````=4sinx-xcosx 于是 y(n)=nsin(x+nπ/2)+xsin[x+(n+1)π/2]

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