题目内容
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[主观题]
设二元函数z=x2y+xsiny,则=()A.2xy+sinyB.x2+xcosyC.2xy+xsinyD.x2y+siny
设二元函数z=x2y+xsiny,则=()
A.2xy+siny
B.x2+xcosy
C.2xy+xsiny
D.x2y+siny
答案
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设二元函数z=x2y+xsiny,则=()
A.2xy+siny
B.x2+xcosy
C.2xy+xsiny
D.x2y+siny