题目内容
(请给出正确答案)
[主观题]
设z=sin(xy)+2x2+y,则dz= .
设z=sin(xy)+2x2+y,则dz= .
答案
[ycos(xy)+4x]dx+[xcos(xy)+1]dy
[解析] dz=d[sin(xy)]+d(2x2)+dy
=cos(xy)(ydx+xdy)+4xdx+dy
=[ycos(xy)+4x]dx+[xcos(xy)+1]dy.
[解析] dz=d[sin(xy)]+d(2x2)+dy
=cos(xy)(ydx+xdy)+4xdx+dy
=[ycos(xy)+4x]dx+[xcos(xy)+1]dy.
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