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z=x/根号下x^2+y^2,求全微分
人气:463 ℃ 时间:2020-03-26 12:35:42
解答
zx=[√(x²+y²)-x²/√(x²+y²)]/(x²+y²)
=y²/(x²+y²)^(3/2)
zy=[0×√(x²+y²)-xy/√(x²+y²)]/(x²+y²)
=-xy/(x²+y²)^(3/2)
所以
dz=zxdx+zydy
=y²/(x²+y²)^(3/2)dx-xy/(x²+y²)^(3/2)dy
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