求二阶偏导的小技巧：复用一阶偏导的部分结果

一、题目

已知函数 $f(u, v)$ 具有二阶连续偏导数, 且满足 $4 \frac{\partial^{2} f}{\partial u^{2}}$ $-$ $\frac{\partial^{2} f}{\partial v^{2}}$ $=$ $1$, 若令 $g(x, y)$ $=$ $f\left(x^{2}+\right.$ $\left.y^{2}, x y\right)$, 则 $\frac{\partial^{2} g}{\partial x^{2}}$ $-$ $\frac{\partial^{2} g}{\partial y^{2}}$ $=$ $?$

难度评级：

二、解析

$$
\frac{\partial g}{\partial x} = \textcolor{orange}{2 x} f_{1}^{\prime} + \textcolor{orange}{y} f_{2}^{\prime} \Rightarrow
$$

小 技 巧：上式中中橙色的 $2x$ 和 $y$ 分别是对函数 $f$ 的第一部分和第二部分变量求导得到的，因此，在下面求解 $f$ 的二阶偏导过程中，再次对函数 $f$ 的第一部分和第二部分变量求导时，就可以直接使用这两个值。

$$
\frac{\partial^{2} g}{\partial x^{2}} =
$$

$$
2 f_{1}^{\prime}+2 x \cdot \textcolor{orange}{2 x} f_{11}^{\prime \prime}+2 x \cdot \textcolor{orange}{y} f_{12}^{\prime \prime} + y \cdot \textcolor{orange}{2 x} f_{21}^{\prime \prime}+y \cdot \textcolor{orange}{y} f_{22}^{\prime \prime}
$$

Next

$$
\frac{\partial g}{\partial y} = 2y f^{\prime}_{1} + x f^{\prime}_{2} \Rightarrow
$$

$$ \frac{\partial^{2} g}{\partial y^{2}} =
$$

$$
2 f_{1}^{\prime}+2 y \cdot 2 y f_{11}^{\prime \prime}+2 y \cdot x f_{12}^{\prime \prime} +x \cdot 2 y f_{21}^{\prime \prime}+x \cdot x f_{22}^{\prime \prime}
$$

Next

于是：

$$
\frac{\partial^{2} g}{\partial x^{2}}-\frac{\partial^{2} g}{\partial y^{2}} =
$$

$$
\left(4 x^{2}-4 y^{2}\right) f_{11}^{\prime \prime}+\left(y^{2}-x^{2}\right) f_{22}^{\prime \prime} =
$$

$$
\left(x^{2}-y^{2}\right) 4 f_{11}^{\prime \prime}-\left(x^{2}-y^{2}\right) f_{22}^{\prime \prime} =
$$

$$
\left(x^{2}-y^{2}\right)\left(4 f_{11}^{\prime \prime}-f_{22}^{\prime \prime}\right)
$$

Next

又因为：

$$
4 \frac{\partial^{2} f}{\partial u^{2}}-\frac{\partial^{2} f}{\partial v^{2}}=1 \Rightarrow 4 f_{11}^{\prime \prime}-f_{22}^{\prime \prime}=1
$$

Next

所以：

$$
\frac{\partial^{2} g}{\partial x^{2}}-\frac{\partial^{2} g}{\partial y^{2}}=\left(x^{2}-y^{2}\right) \times 1=x^{2}-y^{2}.
$$

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