z=x/根号下x^2+y^2,求全微分

如题所述

第1个回答  2012-03-20
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本回答被提问者采纳