复合函数求偏导：z=ln(e^x+y) 的偏导数 z(u) 和 z(v)

首先使用链式法则求得：

$\frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial u}$

$\frac{\partial z}{\partial v}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial v}$

其中，

$\frac{\partial z}{\partial x}=\frac{1}{e^x+y}$

$\frac{\partial z}{\partial y}=\frac{1}{e^x+y}$

$\frac{\partial x}{\partial u}=v$

$\frac{\partial x}{\partial v}=u$

$\frac{\partial y}{\partial u}=1$

$\frac{\partial y}{\partial v}=-1$

将上述结果代入原式中，得到：

$\frac{\partial z}{\partial u}=\frac{v}{e^x+y}$

$\frac{\partial z}{\partial v}=\frac{u}{e^x+y}$

将$x=u*v$，$y=u-v$代入，得到最终结果：

$\frac{\partial z}{\partial u}=\frac{v}{e^{u*v}+u-v}$

$\frac{\partial z}{\partial v}=\frac{u}{e^{u*v}+u-v}$

因此，$z(u)$和$z(v)$分别为：

$z(u)=\int \frac{v}{e^{u*v}+u-v}du+C$

$z(v)=\int \frac{u}{e^{u*v}+u-v}dv+C$，其中$C$为常数。