求u=xyzex+y+z的全微分.

admin2018-08-23  19

问题 求u=xyzex+y+z的全微分.

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答案因u’x=(1+x)yzex+y+z,u’y=(1+y)xzex+y+z,u’z=(1+x)xyex+y+z, 故 du=ex+y+z[(1+x)yzdx+(1+y)xzdy+(1+z)xydz].

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