2025年考研数二第01题解析：偏导数、全微分

一、题目

»A«. $\frac{z}{z+1} \left(\mathrm{e}^{-x^{2}} – \mathrm{e}^{ -y^{2}} \right)$

»B«. $\frac{z}{z+1} \left( \mathrm{e}^{-x^{2}} + \mathrm{e}^{-y^{2}} \right)$

»C«. $- \frac{z}{z+1} \left(\mathrm{e}^{-x^{2}} – \mathrm{e}^{-y^{2}} \right)$

»D«. $- \frac{z}{z+1} \left( \mathrm{e}^{-x^{2}} + \mathrm{e}^{-y^{2}} \right)$

二、解析

方法 1：显式求偏导

之所以将该方法称为“显式求偏导”，是因为在求偏导的过程中考虑了 $z$ 是 $x$ 和 $y$ 的函数这一条件。

在 $z + \ln z – \int_{y}^{x} \mathrm{e}^{-t^{2}} \mathrm{~d} t$ $=$ $0$ 等号两端同时对 $x$ 求偏导，可得：

$$
\begin{align}
& \frac{\partial z}{\partial x} + \frac{1}{z} \frac{\partial z}{\partial x} – \mathrm{e}^{-x^{2}} = 0 \notag \\ \notag \\
\Rightarrow \ & \frac{\partial z}{\partial x} \left( 1 + \frac{1}{z} \right) – \mathrm{e}^{-x^{2}} = 0 \notag \\ \notag \\
\Rightarrow \ & \textcolor{yellow}{ \frac{\partial z}{\partial x} = \frac{z}{z + 1} \mathrm{e}^{-x^{2}} } \tag{1}
\end{align}
$$

接着，由 $z + \ln z – \int_{y}^{x} \mathrm{e}^{-t^{2}} \mathrm{~d} t$ $=$ $0$ 可得：

$$
z + \ln z + \int_{x}^{y} \mathrm{e}^{-t^{2}} \mathrm{~d} t = 0 \tag{2}
$$

于是，在上面的 $(1)$ 式等号两端同时对 $y$ 求偏导，可得：

$$
\begin{align}
& \frac{\partial z}{\partial y} + \frac{1}{z} \frac{\partial z}{\partial y} + \mathrm{e}^{- y^{2}} = 0 \notag \\ \notag \\
\Rightarrow \ & \frac{\partial z}{\partial y} \left( 1 + \frac{1}{z} \right) + \mathrm{e}^{- y^{2}} = 0 \notag \\ \notag \\
\Rightarrow \ & \textcolor{yellow}{ \frac{\partial z}{\partial y} = \frac{-z}{z + 1} \mathrm{e}^{-y^{2}} } \tag{3}
\end{align}
$$

于是，联立 $(1)$ 式和 $(3)$ 式，可得：

$$
\textcolor{lightgreen}{
\boldsymbol{
\frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = \frac{z}{z + 1} \left(\mathrm{e}^{- x^{2}} – \mathrm{e}^{-y^{2}} \right)
}
}
$$

方法 2：基于全微分的隐式求偏导

首先，由 $0$ $=$ $z + \ln z – \int_{y}^{x} \mathrm{e}^{-t^{2} } \mathrm{~d} t$, 得：

$$
\begin{aligned}
& \ 0 = \mathrm{d} \left( z + \ln z – \int_{y}^{x} \mathrm{e}^{-t^{2}} \mathrm{~d} t = 0 \right) \\ \\
\textcolor{lightgreen}{ \leadsto } & \ 0 = \mathrm{d} z + \frac{1}{z} \mathrm{~d} z – \mathrm{e}^{-x^{2}} \mathrm{~d} x – \left( – \mathrm{e}^{ -y^{2}} \right) \mathrm{~d} y \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \mathrm{d} z + \frac{1}{z} \mathrm{~d} z = \mathrm{e}^{-x^{2}} \mathrm{~d} x – \mathrm{e}^{-y^{2}} \mathrm{~d} y \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \mathrm{d} z = \frac{z}{z+1} \left( \mathrm{e}^{ -x^{2}} \mathrm{~d} x – \mathrm{e}^{ -y^{2}} \mathrm{~d} y \right) \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \mathrm{d} z = \frac{z}{z+1} \mathrm{e}^{ -x^{2}} \mathrm{~d} x – \frac{z}{z+1} \mathrm{e}^{ -y^{2}} \mathrm{~d} y \\ \\
\end{aligned}
$$

又由全微分的定义可知：

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

于是可知：

$$
\begin{aligned}
\textcolor{lightgreen}{ \boldsymbol{ \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} } }
& = \frac{z \mathrm{e}^{-x^{2}}}{1+z} – \frac{z \mathrm{e}^{-y^{2}}}{1 + z} \\ \\
& = \textcolor{lightgreen}{ \boldsymbol{ \frac{z}{z+1} \left( \mathrm{e}^{-x^{2}} – \mathrm{e}^{-y^{2}} \right) }}
\end{aligned}
$$

方法 3：基于隐函数的隐式求偏导

首先，设 $F(x,y,z)$ $=$ $z + \ln z – \int_{y}^{x} \mathrm{e}^{-t^{2}} \mathrm{~d} t$, 则：

$$
\begin{aligned}
F_{x}^{\prime} & = -\mathrm{e}^{-x^{2}} \\ \\
F_{y}^{\prime} & = \mathrm{e}^{-y^{2}} \\ \\
F_{z}^{\prime} & = 1 + \frac{1}{z}
\end{aligned}
$$

接着，由隐函数求导公式，可知：

$$
\begin{aligned}
\frac{\partial z}{\partial x} & = \frac{-F_{x}^{\prime}}{F_{z}^{\prime}} = \frac{\mathrm{e}^{-x^{2}}}{1+\frac{1}{z}} = \frac{z\mathrm{e}^{-x^{2}}}{1+z} \\ \\
\frac{\partial z}{\partial y} & = \frac{-F_{y}^{\prime}}{F_{z}^{\prime}} = \frac{-\mathrm{e}^{-y^{2}}}{1+\frac{1}{z}} = \frac{-z \mathrm{e}^{-y^{2}}}{1+z}
\end{aligned}
$$

于是：

$$
\begin{aligned}
\textcolor{lightgreen}{\boldsymbol{ \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} }} & = \frac{z\mathrm{e}^{-x^{2}}}{1+z}-\frac{z\mathrm{e}^{-y^{2}}}{1+z} \\ \\
& = \textcolor{lightgreen}{\boldsymbol{ \frac{z}{z+1}\left(\mathrm{e}^{-x^{2}}-\mathrm{e}^{-y^{2}}\right) }}
\end{aligned}
$$

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