复合函数二阶偏导数问题



u(x,y)=\frac{\frac{\partial z}{\partial x} - \frac{\partial z}{\partial y}}{x - y} (1)
z = x^2 + y^2 - \varphi(x + y + z)(2)
求：\frac{\partial u}{\partial x} = ?
解：
\frac{\partial z}{\partial x} = 2x - \varphi'(\frac{\partial z}{\partial x} + 1) (3)
\frac{\partial z}{\partial y} = 2y - \varphi'(\frac{\partial z}{\partial y} + 1)(4)
由(3)、(4)分别解出：
\frac{\partial z}{\partial x} = \frac{2x - \varphi'}{1 + \varphi'} (5)
\frac{\partial z}{\partial y} = \frac{2y - \varphi'}{1 + \varphi'} (6)
将(5)、(6)代入(1)式，得到：
u(x,y) = \frac{\frac{\partial z}{\partial x} - \frac{\partial z}{\partial y}}{x - y}
= \frac{2}{1 + \varphi'}(7)
这就是第二问题的第一步。而
\frac{\partial u}{\partial x} = -\frac{2\varphi''(1 + \frac{\partial z}{\partial x})}{(1 + \varphi')^2}
将（5）式代入，最后得到：
\frac{\partial u}{\partial x} = -\frac{2\varphi''(1 + 2x)}{(1 + \varphi')^3} (8)
这是第二问题的最后一步！