2015年考研数二第05题解析

题目

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

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

$$C. – \frac{1}{2}, 0$$

$$D. 0, – \frac{1}{2}$$

解析

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

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

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

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

$$u = x + y;$$

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

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

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

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

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

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

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

$$x+y=1;$$

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

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

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

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

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

$①+②$ 得：

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

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

$① – ②$ 得：

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

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

$$x + y = u;$$

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

$$y=u-x;$$

$$y=xv.$$

$$u-x=xv \Rightarrow$$

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

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

$$f=x^{2} – y^{2} =$$

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

$$f = \frac{2u^{2}}{1+v} – u^{2}.$$

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

$$\frac{2}{1+v} \cdot 2u – 2u=$$

$$2-2=0.$$

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

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

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

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

EOF