题目
设 $z=\frac{y}{x}f(xy)$, 其中函数 $f$ 可微，则 $\frac{x}{y} \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = ?$
$$
A. 2yf^{‘}(xy)
$$
$$
B. -2yf^{‘}(xy)
$$
$$
C. \frac{2}{x}f(xy)
$$
$$
D. -\frac{2}{x}f(xy)
$$
解析
本题就是考察复合函数求偏导的知识。
为了使本题中的复合函数更明显，我们设 $u=xy$, 则有：
$$
z=\frac{y}{x}f(u).
$$
于是：
$$
\frac{\partial z}{\partial x} =
$$
$$
y(-\frac{1}{x^{2}})f(u) + \frac{y}{x}\frac{\partial f(u)}{\partial u} \frac{\partial u}{\partial x}.
$$
$$
\frac{\partial z}{\partial y} =
$$
$$
\frac{1}{x}f(u) + \frac{y}{x} \frac{\partial f(u)}{\partial u} \frac{\partial u}{\partial y}.
$$
又：
$$
u = xy.
$$
所以：
$$
\frac{\partial u}{\partial x} = y.
$$
$$
\frac{\partial u}{\partial y} = x.
$$
于是：
$$
\frac{\partial z}{\partial x} =
$$
$$
y(-\frac{1}{x^{2}})f(u) + \frac{y}{x}\frac{\partial f(u)}{\partial u} y.
$$
$$
\frac{\partial z}{\partial y} =
$$
$$
\frac{1}{x}f(u) + \frac{y}{x} \frac{\partial f(u)}{\partial u} x.
$$
于是：
$$
\frac{x}{y} \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} =
$$
$$
\frac{x}{y} [y(-\frac{1}{x^{2}})f(u) + \frac{y}{x}\frac{\partial f(u)}{\partial u} y] +
$$
$$
\frac{1}{x}f(u) + \frac{y}{x} \frac{\partial f(u)}{\partial u} x =
$$
$$
-\frac{1}{x} f(u) + \frac{\partial f(u)}{\partial u} y + \frac{1}{x}f(u) + y \frac{\partial f(u)}{\partial u} =
$$
$$
2y \frac{\partial f(u)}{\partial u}. ①
$$
又因为 $f(u)$ 是 $u$ 的函数，所以：
$$
\frac{\partial f(u)}{\partial u} = f^{‘}(u) = f^{‘}(xy).
$$
于是：
$$
① = 2y f^{‘}(xy).
$$
综上可知，正确选项为 $A$.
EOF