Question

z=y/[(x^2+y^2)^(1/2)] 全微分

Answer 1

∂z/∂x=（-xy)[(x^2+y^2)^(-1/2-1)]=（-xy)[(x^2+y^2)^(-3/2)]

∂z/∂y=(x^2+y^2)^(-1/2)+（-y^2)[(x^2+y^2)^(-3/2)]=（x^2)[(x^2+y^2)^(-3/2)]

dz=（-xy)[(x^2+y^2)^(-3/2)]dx+（x^2)[(x^2+y^2)^(-3/2)]dy