z=x/[(x^2+y^2)^1/2] 的全微分dz需要具体过程 谢谢啦
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z=x/[(x^2+y^2)^1/2] 的全微分dz需要具体过程 谢谢啦
z=x/[(x^2+y^2)^1/2] 的全微分dz
需要具体过程 谢谢啦
z=x/[(x^2+y^2)^1/2] 的全微分dz需要具体过程 谢谢啦
因为
偏z/偏x=[1/(x^2+y^2)^(1/2)] - (x^2)(x^2+y^2)^(-3/2)
偏z/偏y=-xy(x^2+y^2)^(-3/2)
所以全微分
dz=[[1/(x^2+y^2)^(1/2)] - (x^2)(x^2+y^2)^(-3/2)]dx+[-xy(x^2+y^2)^(-3/2)]dy
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