22、设z=(y)/(x)f(xy)，其中函数f可微，则(x)/(y)(partial z)/(partial x)+(partial z)/(partial y)=( )A. 2yf'(xy)B. -2yf'(xy)C. (2)/(x)f(xy)D. -(2)/(x)f(xy)

本题考查复合函数求偏导数的知识。解题思路是先根据求导公式和法则分别求出$\frac{\partial z}{\partial x}$与$\frac{\partial z}{\partial y}$，再将其代入$\frac{x}{y}\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}$进行化简计算。

步骤一：求$\frac{\partial z}{\partial x}$

已知$z = \frac{y}{x}f(xy)$，根据乘积的求导法则$(uv)^\prime = u^\prime v + uv^\prime$，这里$u=\frac{y}{x}$，$v = f(xy)$。

• 对$u=\frac{y}{x}=yx^{-1}$关于$x$求偏导数，根据求导公式$(x^n)^\prime=nx^{n - 1}$可得：
  $\frac{\partial u}{\partial x}=y\times(-1)x^{-2}=-\frac{y}{x^2}$
• 对$v = f(xy)$关于$x$求偏导数，令$t = xy$，根据复合函数求导法则$\frac{\partial v}{\partial x}=\frac{df(t)}{dt}\cdot\frac{\partial t}{\partial x}$，可得：
  $\frac{\partial v}{\partial x}=f^\prime(xy)\cdot y = yf^\prime(xy)$
• 根据乘积求导法则可得：
  $\frac{\partial z}{\partial x}=\frac{\partial u}{\partial x}v + u\frac{\partial v}{\partial x}=-\frac{y}{x^2}f(xy)+\frac{y}{x}\cdot yf^\prime(xy)=-\frac{y}{x^2}f(xy)+\frac{y^2}{x}f^\prime(xy)$

步骤二：求$\frac{\partial z}{\partial y}$

同样根据乘积的求导法则$(uv)^\prime = u^\prime v + uv^\prime$，这里$u=\frac{y}{x}$，$v = f(xy)$。

• 对$u=\frac{y}{x}$关于$y$求偏导数，可得：
  $\frac{\partial u}{\partial y}=\frac{1}{x}$
• 对$v = f(xy)$关于$y$求偏导数，令$t = xy$，根据复合函数求导法则$\frac{\partial v}{\partial y}=\frac{df(t)}{dt}\cdot\frac{\partial t}{\partial y}$，可得：
  $\frac{\partial v}{\partial y}=f^\prime(xy)\cdot x = xf^\prime(xy)$
• 根据乘积求导法则可得：
  $\frac{\partial z}{\partial y}=\frac{\partial u}{\partial y}v + u\frac{\partial v}{\partial y}=\frac{1}{x}f(xy)+\frac{y}{x}\cdot xf^\prime(xy)=\frac{1}{x}f(xy)+yf^\prime(xy)$

步骤三：计算$\frac{x}{y}\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}$

将$\frac{\partial z}{\partial x}=-\frac{y}{x^2}f(xy)+\frac{y^2}{x}f^\prime(xy)$和$\frac{\partial z}{\partial y}=\frac{1}{x}f(xy)+yf^\prime(xy)$代入$\frac{x}{y}\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}$可得：
$\begin{align*}&\frac{x}{y}\left(-\frac{y}{x^2}f(xy)+\frac{y^2}{x}f^\prime(xy)\right)+\frac{1}{x}f(xy)+yf^\prime(xy)\\=&-\frac{1}{x}f(xy)+yf^\prime(xy)+\frac{1}{x}f(xy)+yf^\prime(xy)\\=&(-\frac{1}{x}f(xy)+\frac{1}{x}f(xy))+(yf^\prime(xy)+yf^\prime(xy))\\=&2yf^\prime(xy)\end{align*}$