设u=f(z),而z是由方程z=x+yg(z)确定的函数,其中f,g均为可微函数.证明du/dy=g(z)du/dx.
问题描述：
设u=f(z),而z是由方程z=x+yg(z)确定的函数,其中f,g均为可微函数.证明du/dy=g(z)du/dx.
答
z=x+yg(z) => dz/dx=1+yg'(z)dz/dx
=>dz/dx=1/(1-yg'(z))
dz/dy=g(z)+yg'(z)dz/dy
=>dz/dy=g(z)/(1-yg'(z))
du/dy=df/dy=(df/dz)·(dz/dy)
=g(z)(df/dz)/(1-yg'(z))
du/dx=df/dx=(df/dz)·(dz/dx)
=(df/dz)/(1-yg'(z))
∴du/dy=g(z)du/dx