Question

高数小题，用全微分形式不变性怎么解呢？

Answer 1

两边同时微分，得到
F1'·d(x+z/y)+F2'·d(y+z/x)=0

F1'·(dx-z/y²·dy+1/y·dz)
+F2'·(-z/x²·dx+dy+1/x·dz)=0

整理得到：
(F1'·1/y+F2'·1/x)dz
=(F2'·z/x²-F1')dx+(F1'·z/y²-F2')dy
所以，
dz=(F2'·z/x²-F1')/(F1'·1/y+F2'·1/x)·dx
+(F1'·z/y²-F2')/(F1'·1/y+F2'·1/x)·dy
=(F2'·yz-F1'·x²y)/(F1'·x²+F2'·xy)·dx
+(F1'·xz-F2'·xy²)/(F1'·xy+F2'·y²)·dy

根据全微分的叠加原理，
两个偏导数分别为：
zx=(F2'·yz-F1'·x²y)/(F1'·x²+F2'·xy)
zy=(F1'·xz-F2'·xy²)/(F1'·xy+F2'·y²)