Question

复合函数 求偏导数 是否是链式法则 求详解

Answer 1

可以这样来：
令u=x+y+z, v=x+y, w=x
则f(u, x, v)=0
两边对x求偏导：∂f/∂u*∂u/∂x+∂f/∂w*∂w/∂x+∂f/∂v*∂v/∂x=0
∂f/∂u*(1+z'x)+∂f/∂w+∂f/∂v=0
得：z'x=-(∂f/∂v+∂f/∂w) /(∂f/∂u)-1=-(f'v+f'w)/f'u-1
同理对y求偏导：∂f/∂u*∂u/∂y+∂f/∂v*∂v/∂y=0
∂f/∂u*(1+z'y)+∂f/∂v=0
得：z'y=-(∂f/∂v)/(∂f/∂u)-1=-f'v/f'u-1