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