📝 题目
4．设 $\frac{x}{z}=\ln \frac{z}{y}$ ，求 $\frac{\partial z}{\partial x}$ 及 $\frac{\partial z}{\partial y}$ ．
💡 答案与解析
[AI解答]
[AI解答] 已知方程 $$ \frac{x}{z} = \ln \frac{z}{y} $$ 首先将其改写为隐函数形式： $$ \frac{x}{z} - \ln \frac{z}{y} = 0 $$ 令 $$ F(x, y, z) = \frac{x}{z} - \ln \frac{z}{y} $$ 则原方程即 $F(x, y, z) = 0$。
我们分别计算偏导数：
先对 $x$ 求偏导（将 $y,z$ 视为常数）： $$ \frac{\partial F}{\partial x} = \frac{1}{z} $$
对 $y$ 求偏导： $$ \frac{\partial F}{\partial y} = -\frac{1}{z/y} \cdot \frac{\partial}{\partial y}\left(\frac{z}{y}\right) = -\frac{y}{z} \cdot \left(-\frac{z}{y^2}\right) = \frac{1}{y} $$ （注意 $\ln(z/y) = \ln z - \ln y$，直接对 $y$ 求导得 $-\frac{1}{y}$，结果一致。）
对 $z$ 求偏导： $$ \frac{\partial F}{\partial z} = \frac{\partial}{\partial z}\left( \frac{x}{z} \right) - \frac{\partial}{\partial z}\left( \ln \frac{z}{y} \right) = -\frac{x}{z^2} - \frac{1}{z/y} \cdot \frac{1}{y} = -\frac{x}{z^2} - \frac{1}{z} $$
由隐函数求导公式： $$ \frac{\partial z}{\partial x} = -\frac{\displaystyle\frac{\partial F}{\partial x}}{\displaystyle\frac{\partial F}{\partial z}} = -\frac{\frac{1}{z}}{-\frac{x}{z^2} - \frac{1}{z}} $$ 化简分母： $$ -\frac{x}{z^2} - \frac{1}{z} = -\frac{x + z}{z^2} $$ 因此 $$ \frac{\partial z}{\partial x} = -\frac{\frac{1}{z}}{-\frac{x+z}{z^2}} = \frac{\frac{1}{z}}{\frac{x+z}{z^2}} = \frac{z}{x+z} $$
同理， $$ \frac{\partial z}{\partial y} = -\frac{\displaystyle\frac{\partial F}{\partial y}}{\displaystyle\frac{\partial F}{\partial z}} = -\frac{\frac{1}{y}}{-\frac{x+z}{z^2}} = \frac{\frac{1}{y}}{\frac{x+z}{z^2}} = \frac{z^2}{y(x+z)} $$
因此最终结果为： $$ \boxed{\frac{\partial z}{\partial x} = \frac{z}{x+z},\quad \frac{\partial z}{\partial y} = \frac{z^2}{y(x+z)}} $$
难度：★★☆☆☆