方企勤 第五章 多元函数微分学 第3题
📝 题目
例 3 设
$$ f\left( {x,y}\right) = \left\{ \begin{array}{ll} {xy}\sin \frac{1}{\sqrt{{x}^{2} + {y}^{2}}}, & {x}^{2} + {y}^{2} \neq 0, \\ 0, & {x}^{2} + {y}^{2} = 0. \end{array}\right. $$
求证:
(1) ${f}_{x}^{\prime }\left( {0,0}\right) ,{f}_{y}^{\prime }\left( {0,0}\right)$ 存在;
(2) ${f}_{x}^{\prime }\left( {x,y}\right)$ 与 ${f}_{y}^{\prime }\left( {x,y}\right)$ 在(0,0)点不连续;
(3) $f\left( {x,y}\right)$ 在(0,0)点可微.
💡 答案解析
证 (1) 因 $f\left( {x,0}\right) \equiv 0$ ,所以 ${f}_{x}^{\prime }\left( {0,0}\right) = 0$ ; 同样因 $f\left( {0,y}\right) \equiv 0$ , 得 ${f}_{y}^{\prime }\left( {0,0}\right) = 0$ .
(2)容易求出
$$ {f}_{x}^{\prime }\left( {x,y}\right) = \left\{ \begin{array}{ll} y\sin \frac{1}{\sqrt{{x}^{2} + {y}^{2}}} - \frac{y{x}^{2}}{{\left( {x}^{2} + {y}^{2}\right) }^{3/2}} \times \cos \frac{1}{\sqrt{{x}^{2} + {y}^{2}}}, & {x}^{2} + {y}^{2} \neq 0, \\ 0, & {x}^{2} + {y}^{2} = 0. \end{array}\right. $$
令 $y = x$ ,
$$ {f}_{x}^{\prime }\left( {x,x}\right) = x\sin \frac{1}{\sqrt{2}x} - \frac{1}{2\sqrt{2}}\cos \frac{1}{\sqrt{2}x} \rightarrow 0\;\left( {x \rightarrow 0}\right) , $$
故 ${f}_{x}^{\prime }\left( {x,y}\right)$ 在(0,0)点不连续. 同理可知
$$ {f}_{y}^{\prime }\left( {x,y}\right) = \left\{ \begin{array}{ll} x\sin \frac{1}{\sqrt{{x}^{2} + {y}^{2}}} - \frac{x{y}^{2}}{{\left( {x}^{2} + {y}^{2}\right) }^{3/2}} \times \cos \frac{1}{\sqrt{{x}^{2} + {y}^{2}}}, & {x}^{2} + {y}^{2} \neq 0, \\ 0, & {x}^{2} + {y}^{2} = 0 \end{array}\right. $$
在(0,0)点不连续.
(3) 由于 $\frac{y}{\sqrt{{x}^{2} + {y}^{2}}}\sin \frac{1}{\sqrt{{x}^{2} + {y}^{2}}}\left( {{x}^{2} + {y}^{2} \rightarrow 0}\right)$ 是有界变量,当 ${x}^{2} + {y}^{2} \rightarrow 0$ 时, $x$ 是无穷小量,所以
$$ f\left( {x,y}\right) - f\left( {0,0}\right) = 0 \cdot x + 0 \cdot y + o\left( \sqrt{{x}^{2} + {y}^{2}}\right) , $$
按微分定义,函数 $f$ 在(0,0)点可微,且 $\mathrm{d}f\left( {0,0}\right) = \left( {0,0}\right)$ 或 $\mathrm{d}f\left( {0,0}\right)$ $= 0 \cdot \mathrm{d}x + 0 \cdot \mathrm{d}y$ . 可见偏导数连续是可微的充分条件,不是必要条件.