2013年考研数二第05题解析

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


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