同济高数 第9章 第9-4-2题

[AI解答]

[AI解答]

已知：
$$
z = u^2 \ln v, \quad u = \frac{x}{y}, \quad v = 3x - 2y.
$$

先求 $\displaystyle \frac{\partial z}{\partial x}$，根据链式法则：
$$
\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x}.
$$

计算各偏导数：
$$
\frac{\partial z}{\partial u} = 2u \ln v, \quad \frac{\partial z}{\partial v} = u^2 \cdot \frac{1}{v}.
$$
$$
\frac{\partial u}{\partial x} = \frac{1}{y}, \quad \frac{\partial v}{\partial x} = 3.
$$

代入：
$$
\frac{\partial z}{\partial x} = (2u \ln v) \cdot \frac{1}{y} + \frac{u^2}{v} \cdot 3.
$$

将 $\displaystyle u = \frac{x}{y}$，$v = 3x - 2y$ 代回：
$$
\frac{\partial z}{\partial x} = \frac{2x}{y^2} \ln(3x - 2y) + \frac{3x^2}{y^2(3x - 2y)}.
$$

再求 $\displaystyle \frac{\partial z}{\partial y}$：
$$
\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y}.
$$

其中：
$$
\frac{\partial u}{\partial y} = -\frac{x}{y^2}, \quad \frac{\partial v}{\partial y} = -2.
$$

代入：
$$
\frac{\partial z}{\partial y} = (2u \ln v) \cdot \left(-\frac{x}{y^2}\right) + \frac{u^2}{v} \cdot (-2).
$$

代回 $u, v$：
$$
\frac{\partial z}{\partial y} = -\frac{2x^2}{y^3} \ln(3x - 2y) - \frac{2x^2}{y^2(3x - 2y)}.
$$

因此最终结果为：
$$
\boxed{\frac{\partial z}{\partial x} = \frac{2x}{y^2} \ln(3x - 2y) + \frac{3x^2}{y^2(3x - 2y)}},
$$
$$
\boxed{\frac{\partial z}{\partial y} = -\frac{2x^2}{y^3} \ln(3x - 2y) - \frac{2x^2}{y^2(3x - 2y)}}.
$$