分段函数的偏导数要分段求解

一、题目

已知，函数 $f(x, y)$ $=$ $\begin{cases} \dfrac{xy}{x^{2}+y^{2}}, & (x, y) \neq (0, 0), \\ 0, & (x, y)=(0, 0), \end{cases}$, 求 $f_{x}(x, y)$ 和 $f_{y}(x, y)$.

二、解析

§2.1 当 $(x, y) \neq (0, 0)$ 时

根据区间上偏导数的求解公式，有：

$$
\begin{aligned}
& f_{x}(x, y) = \frac{y(x^{2}+y^{2}) – xy \cdot 2x}{(x^{2} + y^{2})^{2}} = \frac{y^{3} – x^{2}y}{(x^{2} + y^{2})^{2}} \\ \\
& f_{y}(x, y) = \frac{x(x^{2}+y^{2}) – xy \cdot 2y}{(x^{2} + y^{2})^{2}} = \frac{x^{3} – xy^{2}}{(x^{2} + y^{2})^{2}}
\end{aligned}
$$

§2.2 当 $(x, y) = (0, 0)$ 时

在点 $(0, 0)$ 处，根据一点处偏导数的定义，有：

$$
\begin{aligned}
& f_{x}(0, 0) = \lim_{\Delta x \rightarrow 0}\frac{f(0 + \Delta x, 0) – f(0, 0)}{\Delta x} = 0 \\ \\
& f_{y}(0, 0) = \lim_{\Delta y \rightarrow 0}\frac{f(0, 0 + \Delta y) – f(0, 0)}{\Delta y} = 0
\end{aligned}
$$

综上可知：

$$
\textcolor{lightgreen}{
\begin{aligned}
& f_{x}(x, y) = \begin{cases}
\frac{y^{3} – x^{2}y}{(x^{2} + y^{2})^{2}}, & (x, y) \neq (0, 0) \\
0, & (x, y) = (0, 0)
\end{cases} \\ \\ \\
& f_{y}(x, y) = \begin{cases}
\frac{x^{3} – xy^{2}}{(x^{2} + y^{2})^{2}}, & (x, y) \neq (0, 0) \\
0, & (x, y) = (0, 0)
\end{cases}
\end{aligned}
}
$$

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