2026年考研数二第03题解析：全导数、偏导数、全微分、隐函数求导

一、题目

二、解析

解法一：全微分法

根据全微分计算的不变性，由方程 $x – az$ $=$ $\mathrm{e}^{y+az}$, 可得：

$$
\begin{aligned}
& \mathrm{d} x – \mathrm{d} \left( az \right) = \mathrm{d} \left( \mathrm{e}^{y+az} \right) \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \mathrm{d} x – a \mathrm{~d}z = \mathrm{e}^{y+az} \mathrm{~d} \left( y + az \right) \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \mathrm{d} x – a \mathrm{~d}z = \mathrm{e}^{y+az} \left( \mathrm{d}y + a\mathrm{d}z \right) \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \left(a \mathrm{e}^{y+az} + a\right)\mathrm{~d}z = \mathrm{d}x – \left(\mathrm{e}^{y+az}\right)\mathrm{~d}y \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \mathrm{d}z = \frac{1}{a}\left[\frac{1}{\mathrm{e}^{y+az}+1}\mathrm{~d}x + \frac{-\mathrm{e}^{y+az}}{\mathrm{e}^{y+az}+1}\mathrm{~d}y\right]
\end{aligned}
$$

又由二元函数全微分的定义可知：

$$
\mathrm{d} z = \frac{\partial z}{\partial x} \mathrm{~d} x + \frac{\partial z}{\partial y} \mathrm{~d} y
$$

于是，有：

$$
\begin{aligned}
\frac{\partial z}{\partial x} & = \frac{1}{a} \cdot \frac{1}{\mathrm{e}^{y+az}+1} \\ \\
\frac{\partial z}{\partial y} & = \frac{1}{a}\left(\frac{-\mathrm{e}^{y+az}}{\mathrm{e}^{y+az}+1}\right)
\end{aligned}
$$

综上可知：

$$
\begin{aligned}
\textcolor{lightgreen}{ \frac{\partial z}{\partial x} – \frac{\partial z}{\partial y} } & = \frac{1}{a} \left( \frac{1 + \mathrm{e}^{y+az}}{\mathrm{e}^{y+az}+1} \right) \\ \\
& = \textcolor{lightgreen}{ \frac{1}{a} }
\end{aligned}
$$

综上可知，本 题 应 选 A

解法二：隐函数法

首先，构造如下三元函数：

$$
F(x,y,z) = x – az – \mathrm{e}^{y + az}
$$

在函数 $F(x,y,z)$ 中对自变量 $x$, $y$, $z$ 分别求偏导数，得：

$$
\begin{aligned}
F^{\prime}_{x} & = 1 \\
F^{\prime}_{y} & = -\mathrm{e}^{y+az} \\
F^{\prime}_{z} & = -a – a\mathrm{e}^{y+az} = -a\left[1+\mathrm{e}^{y+az}\right]
\end{aligned}
$$

于是，根据隐函数求导运算公式，可知：

$$
\begin{aligned}
\frac{\partial z}{\partial x} & = \textcolor{orangered}{-} \frac{F ^{\prime} _{x}}{F ^{\prime} _{z}} = \textcolor{orangered}{-} \frac{1}{\textcolor{orangered}{-} a\left[1+\mathrm{e}^{y+az}\right]} = \frac{1}{a\left[1+\mathrm{e}^{y+az} \right]} \\ \\
\frac{\partial z}{\partial y} & = \textcolor{orangered}{-} \frac{F ^{\prime} _{y}}{F ^{\prime} _{z}} = \textcolor{orangered}{-} \frac{\textcolor{orangered}{-} \mathrm{e}^{y+az}}{\textcolor{orangered}{-} a\left[1+\mathrm{e}^{y+az}\right]} = \textcolor{orangered}{-} \frac{\mathrm{e}^{y+az}}{a\left[1+\mathrm{e}^{y+az}\right]}
\end{aligned}
$$

于是可得：

$$
\begin{aligned}
\textcolor{lightgreen}{ \frac{\partial z}{\partial x} – \frac{\partial z}{\partial y} } & = \frac{1}{a\left[1+\mathrm{e}^{y+az} \right]} + \frac{\mathrm{e}^{y+az}}{a\left[1+\mathrm{e}^{y+az}\right]} \\ \\
& = \frac{1 + \mathrm{e}^{y+az}}{a\left[1+\mathrm{e}^{y+az}\right]} \\ \\
& = \textcolor{lightgreen}{ \frac{1}{a} }
\end{aligned}
$$

综上可知，本 题 应 选 A

解法三：直接求偏导数

在方程 $x – az$ $=$ $\mathrm{e}^{y+az}$ 等号两边对 $x$ 求偏导数，得：

$$
\begin{aligned}
& 1-a\frac{\partial z}{\partial x} = a\mathrm{e}^{y+az}\frac{\partial z}{\partial x} \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \frac{\partial z}{\partial x}=\frac{1}{a \left( 1+\mathrm{e}^{y+az} \right)}
\end{aligned}
$$

在方程 $x – az$ $=$ $\mathrm{e}^{y+az}$ 等号两边对 $y$ 求偏导数，得：

$$
\begin{aligned}
& -a\frac{\partial z}{\partial y}=\mathrm{e}^{y+az} \left( 1+a\frac{\partial z}{\partial y} \right) \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \frac{\partial z}{\partial y}=\frac{-\mathrm{e}^{y+az}}{a \left( 1+\mathrm{e}^{y+az} \right)}
\end{aligned}
$$

于是：

$$
\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}=\frac{1}{a}
$$

综上可知，本 题 应 选 A

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