12.设 =(e)^ucos v ,=(e)^usin v =uv. 试求 dfrac (partial z)(partial x) 和 dfrac (partial z)(partial y).

本题考查复合函数求偏导数的知识。解题思路是利用复合函数求偏导的链式法则，先找出$z$关于$u$、$v$的偏导数，以及$u$、$v$关于$x$、$y$的偏导数，再根据链式法则计算$\frac{\partial z}{\partial x}$和$\frac{\partial z}{\partial y}$。

步骤一：求$\frac{\partial z}{\partial u}$和$\frac{\partial z}{\partial v}$

已知$z = uv$，对$z$关于$u$求偏导数，将$v$看作常数，根据求导公式$(X^n)^\prime=nX^{n - 1}$可得：
$\frac{\partial z}{\partial u}=v$
对$z$关于$v$求偏导数，将$u$看作常数，可得：
$\frac{\partial z}{\partial v}=u$

步骤二：求$\frac{\partial u}{\partial x}$、$\frac{\partial v}{\partial x}$、$\frac{\partial u}{\partial y}$和$\frac{\partial v}{\partial y}$

由$x = e^u\cos v$，$y = e^u\sin v$，将两式分别对$x$求偏导数：

• 对$x = e^u\cos v$关于$x$求偏导数，等式左边$\frac{\partial x}{\partial x}=1$，等式右边根据乘积的求导法则$(uv)^\prime=u^\prime v+uv^\prime$可得：
  $\frac{\partial x}{\partial x}=\frac{\partial (e^u\cos v)}{\partial x}=e^u\cos v\frac{\partial u}{\partial x}-e^u\sin v\frac{\partial v}{\partial x}$
  即$1 = e^u\cos v\frac{\partial u}{\partial x}-e^u\sin v\frac{\partial v}{\partial x}$ ①
• 对$y = e^u\sin v$关于$x$求偏导数，等式左边$\frac{\partial y}{\partial x}=0$，等式右边根据乘积的求导法则可得：
  $\frac{\partial y}{\partial x}=\frac{\partial (e^u\sin v)}{\partial x}=e^u\sin v\frac{\partial u}{\partial x}+e^u\cos v\frac{\partial v}{\partial x}$
  即$0 = e^u\sin v\frac{\partial u}{\partial x}+e^u\cos v\frac{\partial v}{\partial x}$ ②
  联立①②组成方程组$\begin{cases}e^u\cos v\frac{\partial u}{\partial x}-e^u\sin v\frac{\partial v}{\partial x}=1\\e^u\sin v\frac{\partial u}{\partial x}+e^u\cos v\frac{\partial v}{\partial x}=0\end{cases}$，将$e^u$约去，得到$\begin{cases}\cos v\frac{\partial u}{\partial x}-\sin v\frac{\partial v}{\partial x}=e^{-u}\\\sin v\frac{\partial u}{\partial x}+\cos v\frac{\partial v}{\partial x}=0\end{cases}$。
  由②式可得$\frac{\partial v}{\partial x}=-\frac{\sin v}{\cos v}\frac{\partial u}{\partial x}$，将其代入①式可得：
  $\cos v\frac{\partial u}{\partial x}-\sin v(-\frac{\sin v}{\cos v}\frac{\partial u}{\partial x})=e^{-u}$
  $\cos v\frac{\partial u}{\partial x}+\frac{\sin^2 v}{\cos v}\frac{\partial u}{\partial x}=e^{-u}$
  $\frac{\cos^2 v+\sin^2 v}{\cos v}\frac{\partial u}{\partial x}=e^{-u}$
  因为$\cos^2 v+\sin^2 v = 1$，所以$\frac{1}{\cos v}\frac{\partial u}{\partial x}=e^{-u}$，则$\frac{\partial u}{\partial x}=e^{-u}\cos v$。
  将$\frac{\partial u}{\partial x}=e^{-u}\cos v$代入$\frac{\partial v}{\partial x}=-\frac{\sin v}{\cos v}\frac{\partial u}{\partial x}$可得：
  $\frac{\partial v}{\partial x}=-\frac{\sin v}{\cos v}\cdot e^{-u}\cos v=-e^{-u}\sin v$

同理，将两式分别对$y$求偏导数：

• 对$x = e^u\cos v$关于$y$求偏导数，等式左边$\frac{\partial x}{\partial y}=0$，等式右边根据乘积的求导法则可得：
  $\frac{\partial x}{\partial y}=\frac{\partial (e^u\cos v)}{\partial y}=e^u\cos v\frac{\partial u}{\partial y}-e^u\sin v\frac{\partial v}{\partial y}$
  即$0 = e^u\cos v\frac{\partial u}{\partial y}-e^u\sin v\frac{\partial v}{\partial y}$ ③
• 对$y = e^u\sin v$关于$y$求偏导数，等式左边$\frac{\partial y}{\partial y}=1$，等式右边根据乘积的求导法则可得：
  $\frac{\partial y}{\partial y}=\frac{\partial (e^u\sin v)}{\partial y}=e^u\sin v\frac{\partial u}{\partial y}+e^u\cos v\frac{\partial v}{\partial y}$
  即$1 = e^u\sin v\frac{\partial u}{\partial y}+e^u\cos v\frac{\partial v}{\partial y}$ ④
  联立③④组成方程组$\begin{cases}e^u\cos v\frac{\partial u}{\partial y}-e^u\sin v\frac{\partial v}{\partial y}=0\\e^u\sin v\frac{\partial u}{\partial y}+e^u\cos v\frac{\partial v}{\partial y}=1\end{cases}$，将$e^u$约去，得到$\begin{cases}\cos v\frac{\partial u}{\partial y}-\sin v\frac{\partial v}{\partial y}=0\\\sin v\frac{\partial u}{\partial y}+\cos v\frac{\partial v}{\partial y}=e^{-u}\end{cases}$。
  由③式可得$\frac{\partial v}{\partial y}=\frac{\cos v}{\sin v}\frac{\partial u}{\partial y}$，将其代入④式可得：
  $\sin v\frac{\partial u}{\partial y}+\cos v(\frac{\cos v}{\sin v}\frac{\partial u}{\partial y})=e^{-u}$
  $\sin v\frac{\partial u}{\partial y}+\frac{\cos^2 v}{\sin v}\frac{\partial u}{\partial y}=e^{-u}$
  $\frac{\sin^2 v+\cos^2 v}{\sin v}\frac{\partial u}{\partial y}=e^{-u}$
  因为$\cos^2 v+\sin^2 v = 1$，所以$\frac{1}{\sin v}\frac{\partial u}{\partial y}=e^{-u}$，则$\frac{\partial u}{\partial y}=e^{-u}\sin v$。
  将$\frac{\partial u}{\partial y}=e^{-u}\sin v$代入$\frac{\partial v}{\partial y}=\frac{\cos v}{\sin v}\frac{\partial u}{\partial y}$可得：
  $\frac{\partial v}{\partial y}=\frac{\cos v}{\sin v}\cdot e^{-u}\sin v=e^{-u}\cos v$

步骤三：根据链式法则求$\frac{\partial z}{\partial x}$和$\frac{\partial z}{\partial y}$

根据链式法则$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x}$，将$\frac{\partial z}{\partial u}=v$，$\frac{\partial z}{\partial v}=u$，$\frac{\partial u}{\partial x}=e^{-u}\cos v$，$\frac{\partial v}{\partial x}=-e^{-u}\sin v$代入可得：
$\frac{\partial z}{\partial x}=v\cdot e^{-u}\cos v+u\cdot (-e^{-u}\sin v)=(v\cos v - u\sin v)e^{-u}$
根据链式法则$\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y}$，将$\frac{\partial z}{\partial u}=v$，$\frac{\partial z}{\partial v}=u$，$\frac{\partial u}{\partial y}=e^{-u}\sin v$，$\frac{\partial v}{\partial y}=e^{-u}\cos v$代入可得：
$\frac{\partial z}{\partial y}=v\cdot e^{-u}\sin v+u\cdot e^{-u}\cos v=(u\cos v + v\sin v)e^{-u}$