📝 题目
＊19．设 $f(x, y)=\left\{\begin{array}{ll}x y \frac{x^{2}-y^{2}}{x^{2}+y^{2}}, & (x, y) \neq(0,0), \\ 0, & (x, y)=(0,0),\end{array}\right.$ 试求 $f_{x y}(0,0)$ 及 $f_{y x}(0,0)$ 。
💡 答案与解析
[AI解答]
[AI解答]
已知函数 $$ f(x,y) = \begin{cases} \displaystyle{xy\frac{x^{2}-y^{2}}{x^{2}+y^{2}}}, & (x,y)\neq(0,0),\\ 0, & (x,y)=(0,0). \end{cases} $$
我们首先求一阶偏导数在原点处的值，再求混合偏导。
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**第一步：求 $f_x(0,0)$ 和 $f_y(0,0)$**
由偏导数定义： $$ f_x(0,0) = \lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h} = \lim_{h\to 0}\frac{0-0}{h}=0. $$ 同理， $$ f_y(0,0) = \lim_{k\to 0}\frac{f(0,k)-f(0,0)}{k}=0. $$
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**第二步：求 $f_{xy}(0,0)$**
混合偏导定义为： $$ f_{xy}(0,0) = \lim_{k\to 0}\frac{f_x(0,k)-f_x(0,0)}{k}. $$ 为此，先求 $f_x(0,k)$。当 $k\neq 0$ 时，点 $(0,k)\neq(0,0)$，故 $$ f_x(0,k) = \left.\frac{\partial}{\partial x}\left(xy\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right)\right|_{(0,k)}. $$ 先对 $x$ 求偏导（视 $y$ 为常数）： $$ \frac{\partial}{\partial x}\left(xy\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right) = y\cdot\frac{x^{2}-y^{2}}{x^{2}+y^{2}} + xy\cdot\frac{\partial}{\partial x}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right). $$ 而 $$ \frac{\partial}{\partial x}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right) = \frac{2x(x^{2}+y^{2}) - (x^{2}-y^{2})\cdot 2x}{(x^{2}+y^{2})^{2}} = \frac{4xy^{2}}{(x^{2}+y^{2})^{2}}. $$ 因此 $$ f_x(x,y) = y\frac{x^{2}-y^{2}}{x^{2}+y^{2}} + xy\cdot\frac{4xy^{2}}{(x^{2}+y^{2})^{2}} = y\frac{x^{2}-y^{2}}{x^{2}+y^{2}} + \frac{4x^{2}y^{3}}{(x^{2}+y^{2})^{2}}. $$ 代入 $x=0$，得 $$ f_x(0,k) = k\cdot\frac{0 - k^{2}}{0+k^{2}} + 0 = k\cdot(-1) = -k. $$ 于是 $$ f_{xy}(0,0) = \lim_{k\to 0}\frac{-k - 0}{k} = -1. $$
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**第三步：求 $f_{yx}(0,0)$**
类似地， $$ f_{yx}(0,0) = \lim_{h\to 0}\frac{f_y(h,0)-f_y(0,0)}{h}. $$ 先求 $f_y(h,0)$。当 $h\neq 0$ 时， $$ f_y(x,y) = \frac{\partial}{\partial y}\left(xy\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right). $$ 由对称性，交换 $x$ 与 $y$ 并注意符号：原函数关于 $x,y$ 是奇对称？我们直接计算： $$ f_y(x,y) = x\frac{x^{2}-y^{2}}{x^{2}+y^{2}} + xy\cdot\frac{\partial}{\partial y}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right). $$ 而 $$ \frac{\partial}{\partial y}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right) = \frac{-2y(x^{2}+y^{2}) - (x^{2}-y^{2})\cdot 2y}{(x^{2}+y^{2})^{2}} = \frac{-4x^{2}y}{(x^{2}+y^{2})^{2}}. $$ 所以 $$ f_y(x,y) = x\frac{x^{2}-y^{2}}{x^{2}+y^{2}} - \frac{4x^{3}y^{2}}{(x^{2}+y^{2})^{2}}. $$ 代入 $y=0$，得 $$ f_y(h,0) = h\cdot\frac{h^{2}-0}{h^{2}+0} - 0 = h. $$ 于是 $$ f_{yx}(0,0) = \lim_{h\to 0}\frac{h - 0}{h} = 1. $$
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**结论**： $$ \boxed{f_{xy}(0,0) = -1,\quad f_{yx}(0,0)=1}. $$ 可见混合偏导不相等，说明该函数在原点处二阶混合偏导不连续。
难度：★★★★☆