2015年考研数二第05题解析

题目

设函数 $f(u,v)$ 满足 $f(x + y, \frac{y}{x}) = x^{2} – y^{2}$, 则 $\frac{\partial f}{\partial u} |_{u=1,v=1}$ 和 $\frac{\partial f}{\partial v} |_{u=1,v=1}$ 依次是 $?$

$$
A. \frac{1}{2}, 0
$$

$$
B. 0, \frac{1}{2}
$$

$$
C. – \frac{1}{2}, 0
$$

$$
D. 0, – \frac{1}{2}
$$

解析

方法一:

本题考察的是复合函数求偏导,因此,可以直接使用复合函数求偏导的公式计算。

已知:

$$
\frac{\partial f}{\partial x} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial x};
$$

$$
\frac{\partial f}{\partial y} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial y}.
$$

又(把 $f$ 直接看成 $x$ 和 $y$ 的函数而不是复合函数):

$$
\frac{\partial f}{\partial x} = (x^{2} – y^{2})_{x}^{‘} = 2x;
$$

$$
\frac{\partial f}{\partial y} = (x^{2} – y^{2})_{y}^{‘} = -2y.
$$

令:

$$
u = x + y;
$$

$$
v = \frac{y}{x}.
$$

则:

$$
\frac{\partial u}{\partial x} = 1;
$$

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

$$
\frac{\partial v}{\partial x} = – y x^{-2};
$$

$$
\frac{\partial v}{\partial y} = \frac{1}{x}.
$$

于是有:

$$
2x = \frac{\partial f}{\partial u} + \frac{\partial f}{\partial v} (- y x^{-2});
$$

$$
-2y = \frac{\partial f}{\partial u} + \frac{\partial f}{\partial v} (\frac{1}{x}).
$$

又,当 $u = 1$, $v = 1$ 时,有:

$$
x+y=1;
$$

$$
\frac{y}{x} = 1.
$$

即:

$$
x=\frac{1}{2};
$$

$$
y=\frac{1}{2}.
$$

于是有:

$$
1 = \frac{\partial f}{\partial u} – 2 \frac{\partial f}{\partial v}; ①
$$

$$
-1 = \frac{\partial f}{\partial u} + 2 \frac{\partial f}{\partial v}. ②
$$

$①+②$ 得:

$$
0=2 \frac{\partial f}{\partial u} \Rightarrow
$$

$$
\frac{\partial f}{\partial u} = 0.
$$

$① – ②$ 得:

$$
2=-4\frac{\partial f}{\partial v} \Rightarrow
$$

$$
\frac{\partial f}{\partial v} = – \frac{1}{2}.
$$

方法二:

从题目中给出的函数 $f$ 的形式可以看出,$f$ 是关于 $x$ 和 $y$ 的复合函数。如果,我们能用 $u$ 和 $v$ 表示 $x$ 和 $y$, 那么,就可以把 $f$ 变成关于 $u$ 和 $v$ 的非复合函数,从而 $\frac{\partial f}{\partial u}$ 和 $\frac{\partial f}{\partial v}$ 就是对非复合函数求偏导。

已知:

$$
x + y = u;
$$

$$
\frac{y}{x} = v.
$$

于是:

$$
y=u-x;
$$

$$
y=xv.
$$

于是:

$$
u-x=xv \Rightarrow
$$

$$
x = \frac{u}{1+v}.
$$

进而:

$$
y=u-\frac{u}{1+v}.
$$

于是:

$$
f=x^{2} – y^{2} =
$$

$$
(\frac{u}{1+v})^{2} – (u-\frac{u}{1+v})^{2} \Rightarrow
$$

$$
f = \frac{2u^{2}}{1+v} – u^{2}.
$$

于是,当 $u=1$, $v=1$ 时:

$$
\frac{\partial f}{\partial u} =
$$

$$
\frac{2}{1+v} \cdot 2u – 2u=
$$

$$
2-2=0.
$$

$$
\frac{\partial f}{\partial v} =
$$

$$
2u^{2} \times (\frac{1}{1+v})^{‘} =
$$

$$
2u^{2} \times (-1)\frac{1}{(1+v)^{2}} =
$$

$$
2 \times (- \frac{1}{4}) = – \frac{1}{2}.
$$

综上可知,正确选项为 $D$.

EOF