二阶偏导数求导对比：两个变量的三元函数和三个变量的二元函数

题目二：三个变量的二元函数

设 $u$ $=$ $f \left( x+y+z, x^{2} + y^{2} + z^{2} \right)$, 求 $\frac{\partial^{2} u }{\partial x^{2}}$, $\frac{\partial^{2} u}{\partial x \partial y}$, $\frac{\partial^{2} u}{\partial y^{2}}$.

其中，$f$ 具有二阶连续偏导数。

难度评级：

解析二

首先，求解一阶偏导数：

$$
\begin{aligned}
\frac{\partial u}{\partial x} & = f_{1}^{\prime} + 2x f_{2}^{\prime} \\ \\
\frac{\partial u}{\partial y} & = f_{1}^{\prime} + 2y f_{2}^{\prime}
\end{aligned}
$$

于是：

$$
\begin{aligned}
\frac{\partial^{2} u}{\partial x^{2} } \\ \\
= & \ \frac{\partial u}{\partial x} \left( f_{1}^{\prime} + 2x f_{2}^{\prime} \right) \\ \\
= & \ f_{11}^{\prime \prime} + 2x f_{12}^{\prime \prime} + 2 f_{2}^{\prime} + 2x \left( f_{21}^{\prime \prime} + 2x f_{22}^{ \prime \prime } \right) \\ \\
= & \ \textcolor{springgreen}{\boldsymbol{ f_{11}^{\prime \prime} + 4x f_{12}^{\prime \prime} + 2 f_{2}^{\prime} + 4x^{2} f_{22}^{\prime \prime} }}
\end{aligned}
$$

于是：

$$
\begin{aligned}
\frac{\partial^{2} u }{\partial x \partial y} \\ \\
= & \ \frac{\partial u}{\partial y} \left( f_{1}^{\prime} + 2y f_{2}^{\prime} \right) \\ \\
= & \ f_{11}^{\prime \prime} + 2y f_{12}^{\prime \prime} + 2x \left( f_{21}^{\prime \prime} + 2y f_{22}^{\prime \prime} \right) \\ \\
= & \ \textcolor{springgreen}{\boldsymbol{ f_{11}^{\prime \prime} + 2 (x+y) f_{12}^{\prime \prime} + 4x y f_{22}^{\prime \prime} }}
\end{aligned}
$$

于是：

$$
\begin{aligned}
\frac{\partial^{2} u}{\partial y^{2}} \\ \\
= & \ \frac{\partial u}{\partial y} \left( f_{1}^{\prime} + 2y f_{2}^{\prime} \right) \\ \\
= & \ f_{11}^{\prime \prime} + 2y f_{12}^{\prime \prime} + 2 f_{2}^{\prime} + 2y \left( f_{21}^{\prime \prime} + 2y f_{22}^{\prime \prime} \right) \\ \\
= & \ \textcolor{springgreen}{\boldsymbol{ f_{11}^{\prime \prime} + 4y f_{12}^{\prime \prime} + 2 f_{2}^{\prime} + 4y^{2} f_{22}^{\prime \prime} }}
\end{aligned}
$$

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