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求z=exsin(x+y)的全微分.
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【答案】:由于z'x=exsin(x+y)+excos(x+y), z'y=excos(x+y),
所以 dz=z'zdx+z'ydy
=ex[sin(x+y)+cos(x+y)]dx+eycos(x+y)dy.
所以 dz=z'zdx+z'ydy
=ex[sin(x+y)+cos(x+y)]dx+eycos(x+y)dy.
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