已知du(x,y)=[axy3+cos(x+2y)]dx+[-3x2y2+bcos(x+2y)]dy,则 ( )

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问题 已知du(x,y)=[axy3+cos(x+2y)]dx+[-3x2y2+bcos(x+2y)]dy,则    (    )

选项 A、a=2,b=一2
B、a=3,b=2
C、a=2,b=2
D、a=一2,b=2

答案C

解析 由du(x,y)=[axy3+cos(x+2y)]dx+[3x2y2+bcos(37+2y)]dy可知,以上两式分别对y,x求偏导得3axy2-2sin(x+2y)=6xy2-bsin(s+2y),故得a=2,b=2.
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