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设u=ln(x+y^2+z^3),求du
一道偏导全微分的题,
人气:223 ℃ 时间:2020-06-06 00:45:49
解答
u'x=1/(x+y^2+z^3)
u'y=2y/(x+y^2+z^3)
u'z=3z^2/(x+y^2+z^3)
du=u'xdx+u'ydy+u'zdz
=1/(x+y^2+z^3)dx+2y/(x+y^2+z^3)dy+3z^2/(x+y^2+z^3)dz
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