zx=1/(1+(x/y)²) *1/y =y/(x²+y²)
zy=1/(1+(x/y)²) *(-x/y²) =-x/(x²+y²)
所以
dz=y/(x²+y²)dx-x/(x²+y²)dy
设函数z=arctanx/y,求全微分dz
zx=1/(1+(x/y)²) *1/y =y/(x²+y²)
zy=1/(1+(x/y)²) *(-x/y²) =-x/(x²+y²)
所以
dz=y/(x²+y²)dx-x/(x²+y²)dy