设函数z=exy(x2+y一1),则.

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问题 设函数z=exy(x2+y一1),则

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答案exy(x2y+2x+y2一y)

解析 函数z=exy(x2+y一1),则=yexy(x2+y一1)+exy(2x)=exy(x2y+y2一y)+2xexy=exy(x2y+2x+y2一y)
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