📝 题目
2．设 $z=u^{2} \ln v$ ，而 $u=\frac{x}{y}, v=3 x-2 y$ ，求 $\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}$ ．
💡 答案与解析
[AI解答]
[AI解答]
已知： $$ z = u^2 \ln v, \quad u = \frac{x}{y}, \quad v = 3x - 2y. $$
先求 $\frac{\partial z}{\partial x}$，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 $u = \frac{x}{y}$，$v = 3x - 2y$ 代回： $$ \frac{\partial z}{\partial x} = \frac{2x}{y^2} \ln(3x - 2y) + \frac{3x^2}{y^2(3x - 2y)}. $$
再求 $\frac{\partial z}{\partial y}$： $$ \frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y}. $$
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： $$ \boxed{\frac{\partial z}{\partial x} = \frac{2x}{y^2} \ln(3x - 2y) + \frac{3x^2}{y^2(3x - 2y)}}, $$ $$ \boxed{\frac{\partial z}{\partial y} = -\frac{2x^2}{y^3} \ln(3x - 2y) - \frac{2x^2}{y^2(3x - 2y)}}. $$