z^2-xy+z=1, 求 ∂²z/∂x∂y
等式两边对y求偏导,得 2zz'<y>-x+z'<y>=0,得 z'<y>=x/(1+2z),
等式两边对x求偏导,得 2zz'<x>-y+z'<x>=0, 得 z'<x>=y/(1+2z),
进而得 z''<xy> = (1+2z-2yz'<y>)/(1+2z)^2 = [1+2z-2xy/(1+2z)]/(1+2z)^2
= [(1+2z)^2-2xy]/(1+2z)^3.
记 F=x+y-u-v=0, G=x/y-sinu/sinv=0, 其中 u=u(x,y),v=v(x,y).
则 F'<x>=1-u'<x>-v'<x>=0, G'<x>=1/y-(u'<x>cosusinv-v'<x>sinucosv)/(sinv)^2=0,
即 u'<x>+v'<x>=1
u'<x>cosusinv-v'<x>sinucosv=(sinv)^2/y
当 sin(u+v)≠0 时联立解得
u'<x>= [(sinv)^2/y+sinucosv]/sin(u+v),
v'<x>=[-(sinv)^2/y+cosusinv]/sin(u+v).