2021年考研数二第06题解析：全微分、二元复合函数求导

一、题目

»A« $\mathrm{d} x + \mathrm{d} y$

»B« $\mathrm{d} x – \mathrm{d} y$

»C« $\mathrm{d} y$

»D« $-\mathrm{d} y$

难度评级：

二、解析

首先，根据全微分的定义可知：

$$
\textcolor{yellow}{
\mathrm{d} f\left(1, 1\right) = f_{1}^{\prime}\left(1, 1\right) \mathrm{d} x + f_{2}^{\prime}\left(1, 1\right) \mathrm{d} y } \tag{1}
$$

于是可知，要求解 $\mathrm{d} f\left(1, 1\right)$ 的表达式，就要首先求解出 $f_{1}^{\prime}\left(1, 1\right)$ 和 $f_{2}^{\prime}\left(1, 1\right)$.

由于 $f_{1}^{\prime}\left(1, 1\right)$ 和 $f_{2}^{\prime}\left(1, 1\right)$ 相当于两个未知量，就需要至少两个等式才能求解，而刚好题目给了两个不同的式子 $f\left(x + 1, \mathrm{e}^{x}\right) = x \left(x + 1\right)^{2}$ 和 $f\left(x, x^{2}\right) = 2 x^{2} \ln x$, 于是：

$\textcolor{lightgreen}{\blacktriangleright}$ 在方程 $f\left(x + 1, e^{x}\right) = x \left(x + 1\right)^{2}$ 的等号两边对 $x$ 求导得：

$$
\begin{align}
& \ f_{1}^{\prime}\left(x + 1, e^{x}\right) + f_{2}^{\prime}\left(x + 1, \mathrm{e}^{x}\right) \mathrm{e}^{x} = \left(x + 1\right)^{2} + 2 x \left(x + 1\right) \notag \\ \notag \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{x = 0} \notag \\ \notag \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{yellow}{ f_{1}^{\prime}\left(1, 1\right) + f_{2}^{\prime}\left(1, 1\right) = 1 } \tag{2}
\end{align}
$$

$\textcolor{lightgreen}{\blacktriangleright}$ 在方程 $f\left(x, x^{2}\right) = 2 x^{2} \ln x$ 的等号两边对 $x$ 求导得：

$$
\begin{align}
& \ f_{1}^{\prime}\left(x, x^{2}\right) + f_{2}^{\prime}\left(x, x^{2}\right) \cdot 2 x = 4 x \ln x + 2 x^{2} \cdot \frac{1}{x} \notag \\ \notag \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{x = 1} \notag \\ \notag \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{yellow}{ f_{1}^{\prime}\left(1, 1\right) + f_{2}^{\prime}\left(1, 1\right) \cdot 2 = 2 } \tag{3}
\end{align}
$$

综上，联立 $(2)$ 式和 $(3)$ 式，得：

$$
\begin{aligned}
f_{1}^{\prime}\left(1, 1\right) & = 0 \\ \\
f_{2}^{\prime}\left(1, 1\right) & = 1
\end{aligned}
$$

于是，结合前面的 $(1)$ 式，可知：

$$
\begin{aligned}
\textcolor{lightgreen}{ \mathrm{d} f\left(1, 1\right) } & = f_{1}^{\prime}\left(1, 1\right) \mathrm{d} x + f_{2}^{\prime}\left(1, 1\right) \mathrm{d} y \\ \\
& = \textcolor{lightgreen}{ \mathrm{d} y }
\end{aligned}
$$

综上可知，本 题 应 选 C

---