人邮高数 第6章 第6-2-13题
📝 题目
13．设 $z=x \ln (x y)$ ，求 $\displaystyle \frac{\partial^{3} z}{\partial x^{2} \partial y}, \frac{\partial^{3} z}{\partial x \partial y^{2}}$ ．
💡 答案解析
[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：求一阶偏导数** $$ \frac{\partial z}{\partial x} = \ln x + x \cdot \frac{1}{x} + \ln y = \ln x + 1 + \ln y $$ $$ \frac{\partial z}{\partial y} = x \cdot \frac{1}{y} = \frac{x}{y} $$
**第二步：求二阶偏导数** 先求 $\frac{\partial^2 z}{\partial x^2}$： $$ \frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x}(\ln x + 1 + \ln y) = \frac{1}{x} $$ 再求 $\frac{\partial^2 z}{\partial x \partial y}$： $$ \frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial y}(\ln x + 1 + \ln y) = \frac{1}{y} $$ 以及 $\frac{\partial^2 z}{\partial y^2}$： $$ \frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y}\left( \frac{x}{y} \right) = -\frac{x}{y^2} $$
**第三步：求三阶偏导数** (1) 求 $\frac{\partial^3 z}{\partial x^2 \partial y}$： 先对 $x$ 求两次，再对 $y$ 求一次。 由 $\frac{\partial^2 z}{\partial x^2} = \frac{1}{x}$，再对 $y$ 求偏导： $$ \frac{\partial^3 z}{\partial x^2 \partial y} = \frac{\partial}{\partial y}\left( \frac{1}{x} \right) = 0 $$
(2) 求 $\frac{\partial^3 z}{\partial x \partial y^2}$： 先对 $y$ 求两次，再对 $x$ 求一次。 由 $\frac{\partial^2 z}{\partial y^2} = -\frac{x}{y^2}$，再对 $x$ 求偏导： $$ \frac{\partial^3 z}{\partial x \partial y^2} = \frac{\partial}{\partial x}\left( -\frac{x}{y^2} \right) = -\frac{1}{y^2} $$
**最终结果** $$ \boxed{\frac{\partial^{3} z}{\partial x^{2} \partial y} = 0,\quad \frac{\partial^{3} z}{\partial x \partial y^{2}} = -\frac{1}{y^{2}}} $$
难度：★☆☆☆☆