2026年考研数二第14题解析：多元复合函数求导

一、题目

二、解析

首先，由 $\mathrm{d} f(0,0) = \pi \mathrm{d} x + 3 \mathrm{d} y$ 可知：

$$
\begin{aligned}
& \textcolor{gray}{ \frac{\partial f}{\partial x} \bigg|_{(0,0)} = f ^{\prime} _{x}(0,0) = } f ^{\prime} _{1}(0,0) = \pi \\ \\
& \textcolor{gray}{ \frac{\partial f}{\partial y} \bigg|_{(0,0)} = f ^{\prime} _{y}(0,0) = } f ^{\prime} _{2}(0,0) = 3
\end{aligned}
$$

又因为：

$$
\begin{aligned}
& g ^{\prime} (x) = f ^{\prime} _{1} \cdot \frac{1}{x} + f ^{\prime} _{2} \cdot \pi \cdot \cos (\pi x) \\ \\
& g(1) = f(\ln 1, \sin \pi) = f(0, 0)
\end{aligned}
$$

所以：

$$
\begin{aligned}
g ^{\prime} (1) & = \pi \cdot 1 + 3 \cdot \pi \cdot \cos \pi \\ \\
& = \pi + 3 \cdot \pi \cdot (-1) \\ \\
& = -2 \pi
\end{aligned}
$$

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