人邮高数 第6章 第6-3-2题
📝 题目
2．设 $\displaystyle z=u^{2} \ln v, u=\frac{y}{x}, v=2 x-3 y$ ，求 $\displaystyle \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}$ ．
💡 答案解析
[AI解答]
[AI解答]
已知
$$
z = u^{2} \ln v,\quad u = \frac{y}{x},\quad v = 2x - 3y
$$
先求 $\displaystyle \frac{\partial z}{\partial x}$，由链式法则：
$$
\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x}
$$
计算各部分：
$$
\frac{\partial z}{\partial u} = 2u \ln v,\quad
\frac{\partial z}{\partial v} = u^{2} \cdot \frac{1}{v}
$$
$$
\frac{\partial u}{\partial x} = y \cdot \left(-\frac{1}{x^{2}}\right) = -\frac{y}{x^{2}},\quad
\frac{\partial v}{\partial x} = 2
$$
代入得：
$$
\frac{\partial z}{\partial x} = (2u \ln v)\left(-\frac{y}{x^{2}}\right) + \frac{u^{2}}{v} \cdot 2
$$
将 $\displaystyle u = \frac{y}{x}$，$v = 2x - 3y$ 代回：
$$
\frac{\partial z}{\partial x} = 2\cdot\frac{y}{x} \ln(2x-3y) \cdot \left(-\frac{y}{x^{2}}\right) + \frac{(y/x)^{2}}{2x-3y} \cdot 2
$$
化简：
$$
\frac{\partial z}{\partial x} = -\frac{2y^{2}}{x^{3}} \ln(2x-3y) + \frac{2y^{2}}{x^{2}(2x-3y)}
$$
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
$$
\frac{\partial u}{\partial y} = \frac{1}{x},\quad
\frac{\partial v}{\partial y} = -3
$$
代入得：
$$
\frac{\partial z}{\partial y} = (2u \ln v)\cdot\frac{1}{x} + \frac{u^{2}}{v} \cdot (-3)
$$
代回 $u,v$：
$$
\frac{\partial z}{\partial y} = 2\cdot\frac{y}{x} \ln(2x-3y) \cdot \frac{1}{x} - 3\cdot\frac{(y/x)^{2}}{2x-3y}
$$
化简：
$$
\frac{\partial z}{\partial y} = \frac{2y}{x^{2}} \ln(2x-3y) - \frac{3y^{2}}{x^{2}(2x-3y)}
$$
最终结果为：
$$
\boxed{\frac{\partial z}{\partial x} = -\frac{2y^{2}}{x^{3}} \ln(2x-3y) + \frac{2y^{2}}{x^{2}(2x-3y)}}
$$
$$
\boxed{\frac{\partial z}{\partial y} = \frac{2y}{x^{2}} \ln(2x-3y) - \frac{3y^{2}}{x^{2}(2x-3y)}}
$$
难度：★★☆☆☆