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u=1/2x^2-siny+e^xyz求du
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u=1/2x^2-siny+e^xyz求du
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答案和解析
u'(x) = x+yze^(xyz)
u'(y)=-cosy+xze^(xyz)
u'(z)=xye^(xyz)
所以du = [x+yze^(xyz)] dx + [-cosy+xze^(xyz)]dy + xye^(xyz) dz
u'(y)=-cosy+xze^(xyz)
u'(z)=xye^(xyz)
所以du = [x+yze^(xyz)] dx + [-cosy+xze^(xyz)]dy + xye^(xyz) dz
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