📝 题目
9．设 $z=x \ln (x y)$ ，求 $\frac{\partial^{3} z}{\partial x^{2} \partial y}$ 及 $\frac{\partial^{3} z}{\partial x \partial y^{2}}$ ．
💡 答案与解析
[AI解答]
[AI解答] 已知函数 $$ z = x \ln(xy) $$ 首先，将函数改写为便于求导的形式： $$ z = x \left[ \ln x + \ln y \right] = x \ln x + x \ln y $$
**第一步：求 $\frac{\partial z}{\partial x}$** $$ \frac{\partial z}{\partial x} = \ln x + x \cdot \frac{1}{x} + \ln y = \ln x + 1 + \ln y $$
**第二步：求 $\frac{\partial^2 z}{\partial x^2}$** $$ \frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x} (\ln x + 1 + \ln y) = \frac{1}{x} $$
**第三步：求 $\frac{\partial^3 z}{\partial x^2 \partial y}$** 先对 $y$ 求偏导： $$ \frac{\partial^3 z}{\partial x^2 \partial y} = \frac{\partial}{\partial y} \left( \frac{1}{x} \right) = 0 $$ 因此 $$ \boxed{0} $$
**第四步：求 $\frac{\partial^2 z}{\partial x \partial y}$** 由第一步结果： $$ \frac{\partial z}{\partial x} = \ln x + 1 + \ln y $$ 对 $y$ 求偏导： $$ \frac{\partial^2 z}{\partial x \partial y} = \frac{1}{y} $$
**第五步：求 $\frac{\partial^3 z}{\partial x \partial y^2}$** 对 $y$ 再求一次偏导： $$ \frac{\partial^3 z}{\partial x \partial y^2} = \frac{\partial}{\partial y} \left( \frac{1}{y} \right) = -\frac{1}{y^2} $$ 因此 $$ \boxed{-\frac{1}{y^{2}}} $$
**最终答案：** $$ \frac{\partial^{3} z}{\partial x^{2} \partial y} = 0,\quad \frac{\partial^{3} z}{\partial x \partial y^{2}} = -\frac{1}{y^{2}} $$
难度：★☆☆☆☆