# 2013年考研数二第05题解析

## 题目

$$A. 2yf^{‘}(xy)$$

$$B. -2yf^{‘}(xy)$$

$$C. \frac{2}{x}f(xy)$$

$$D. -\frac{2}{x}f(xy)$$

## 解析

$$z=\frac{y}{x}f(u).$$

$$\frac{\partial z}{\partial x} =$$

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

$$\frac{\partial z}{\partial y} =$$

$$\frac{1}{x}f(u) + \frac{y}{x} \frac{\partial f(u)}{\partial u} \frac{\partial u}{\partial y}.$$

$$u = xy.$$

$$\frac{\partial u}{\partial x} = y.$$

$$\frac{\partial u}{\partial y} = x.$$

$$\frac{\partial z}{\partial x} =$$

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

$$\frac{\partial z}{\partial y} =$$

$$\frac{1}{x}f(u) + \frac{y}{x} \frac{\partial f(u)}{\partial u} x.$$

$$\frac{x}{y} \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} =$$

$$\frac{x}{y} [y(-\frac{1}{x^{2}})f(u) + \frac{y}{x}\frac{\partial f(u)}{\partial u} y] +$$

$$\frac{1}{x}f(u) + \frac{y}{x} \frac{\partial f(u)}{\partial u} x =$$

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

$$2y \frac{\partial f(u)}{\partial u}. ①$$

$$\frac{\partial f(u)}{\partial u} = f^{‘}(u) = f^{‘}(xy).$$

$$① = 2y f^{‘}(xy).$$

EOF