设z=ex2+y2sinxy,求

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问题 设z=ex2+y2sinxy,求

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答案[*]=2xex2+y2sinxy+yex2+y2cosxy; [*]=2yex2+y2sinxy+xex2+y2cosxy; [*]=4xyex2+y2sinxy+2x2ex2+y2cosxy+ex2+y2cosxy+2y2ex2+y2cosxy-xyex2+y2sinccy =ex2+y2[3xysinxy+(2x2+2y2+1)cosxy].

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