110 习题 Exercises

1. 函数 $$f(x,y)=\begin{cases} xy\frac{x^2-y^2}{x^2+y^2},&&(x,y)\neq(0,0)\\ 0,&&(x,y)=(0,0) \end{cases}$$ 在 $(0,0)$ 处是连续的。求 $f_{xy}(0,0),f_{yx}(0,0)$。

解： $$\begin{aligned} \frac{\partial f}{\partial x}&=y\frac{x^2-y^2}{x^2+y^2}+xy\frac{2x(x^2+y^2)-2x(x^2-y^2)}{(x^2+y^2)^2}\\ &=y\frac{x^2-y^2}{x^2+y^2}+xy\frac{4xy^2}{(x^2+y^2)^2}\\ &=y\frac{(x^2-y^2)(x^2-y^2)}{(x^2+y^2)^2}+xy\frac{4xy^2}{(x^2+y^2)^2}\\ &=\frac{yx^4-y^5}{(x^2+y^2)^2}+\frac{4x^2y^3}{(x^2+y^2)^2}\\ &=\frac{yx^4-y^5+4x^2y^3}{(x^2+y^2)^2} \end{aligned}$$ $$\begin{aligned} f_x(0,0)&=\lim_{x\to 0}\frac{f(x,0)-f(0,0)}{x}\\ &=\lim_{x\to 0}\frac{0-0}{x}\\ &=0 \end{aligned}$$ $$\begin{aligned} f_{xy}(0,0)&=\lim_{y\to 0}\frac{f_x(0,y)-f_x(0,0)}{y}\\ &=\lim_{y\to 0}\frac{-y-0}{y}\\ &=-1 \end{aligned}$$ $$\begin{aligned} \frac{\partial f}{\partial y}&=x\frac{x^2-y^2}{x^2+y^2}+xy\frac{-2y(x^2+y^2)-2y(x^2-y^2)}{(x^2+y^2)^2}\\ &=x\frac{x^2-y^2}{x^2+y^2}+xy\frac{-4x^2y}{(x^2+y^2)^2}\\ &=x\frac{(x^2-y^2)(x^2-y^2)}{(x^2+y^2)^2}+xy\frac{-4x^2y}{(x^2+y^2)^2}\\ &=\frac{x^5-xy^4}{(x^2+y^2)^2}+\frac{-4x^3y^2}{(x^2+y^2)^2}\\ &=\frac{x^5-xy^4-4x^3y^2}{(x^2+y^2)^2} \end{aligned}$$ $$\begin{aligned} f_y(0,0)&=\lim_{y\to 0}\frac{f(0,y)-f(0,0)}{y}\\ &=\lim_{y\to 0}\frac{0-0}{y}\\ &=0 \end{aligned}$$ $$\begin{aligned} f_{yx}(0,0)&=\lim_{x\to 0}\frac{f_y(x,0)-f_y(0,0)}{x}\\ &=\lim_{x\to 0}\frac{x-0}{x}\\ &=1 \end{aligned}$$