三元函数全微分的计算：比二元多一元

一、题目

已知 $f(x,y,z)$ $=$ $\left( \frac{x}{y} \right)^{\frac{1}{z}}$, 则：

$$
\mathrm{d} f(1,1,1) = ?
$$

难度评级：

二、解析

首先，根据二元函数的全微分，推广可得三元函数的全微分为：

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

接着，我们对题目所给函数两端做变形处理：

$$
\begin{aligned}
& f(x, y, z) = \left( \frac{x}{y} \right)^{\frac{1}{z}} \\ \\
\Rightarrow & \ \textcolor{orangered}{\ln} f(x, y, z) = \textcolor{orangered}{\ln} \left( \frac{x}{y} \right)^{\frac{1}{z}} \\ \\
\Rightarrow & \ \textcolor{orangered}{\ln} f(x, y, z) = \frac{1}{z} \textcolor{orangered}{\ln} \left( \frac{x}{y} \right) \\ \\
\Rightarrow & \ \textcolor{orangered}{\ln} f(x, y, z) = \frac{1}{z} \textcolor{orangered}{\ln} x – \frac{1}{z} \textcolor{orangered}{\ln} y
\end{aligned}
$$

又因为：

$$
f(1,1,1) = \left( \frac{1}{1} \right)^{\frac{1}{1}} = \textcolor{springgreen}{1}
$$

于是：

$$
\begin{aligned}
& \left[ \ln f \right]^{\prime}_{x} = \left[ \frac{1}{z} \ln x – \frac{1}{z} \ln y \right] ^{\prime} _{x} \\ \\
\Rightarrow & \ \frac{f ^{\prime} _{x}}{f} = \frac{1}{x} \\ \\
\Rightarrow & \ \textcolor{gray}{x = 1} \\ \\
\Rightarrow & \ \textcolor{springgreen}{ f ^{\prime} _{x} = \frac{\partial f}{\partial x} = 1 }
\end{aligned}
$$

于是：

$$
\begin{aligned}
& \left[ \ln f \right]^{\prime}_{y} = \left[ \frac{1}{z} \ln x – \frac{1}{z} \ln y \right] ^{\prime} _{y} \\ \\
\Rightarrow & \ \frac{f ^{\prime} _{y}}{f} = \frac{1}{y} \\ \\
\Rightarrow & \ \textcolor{gray}{y = 1} \\ \\
\Rightarrow & \ \textcolor{springgreen}{ f ^{\prime} _{y} = \frac{\partial f}{\partial y} = -1 }
\end{aligned}
$$

于是：

$$
\begin{aligned}
& \left[ \ln f \right]^{\prime}_{z} = \left[ \frac{1}{z} \ln x – \frac{1}{z} \ln y \right] ^{\prime} _{z} \\ \\
\Rightarrow & \ \textcolor{gray}{\begin{cases}
x = 1 \\
y = 1
\end{cases}} \\ \\
\Rightarrow & \ \left[ \ln f \right]^{\prime}_{z} = \left[ \frac{1}{z} \ln 1 – \frac{1}{z} \ln 1 \right] ^{\prime} _{z} \\ \\
\Rightarrow & \ \frac{f ^{\prime} _{z}}{f} = \left[ 0 \right]^{\prime} _{z} \\ \\
\Rightarrow & \ \textcolor{springgreen}{f ^{\prime} _{z} = \frac{\partial f}{\partial z} = 0}
\end{aligned}
$$

综上可知：

$$
\begin{aligned}
& \mathrm{d} f = \frac{\partial f}{\partial x} \mathrm{~d} x + \frac{\partial f}{\partial y} \mathrm{~d} y + \frac{\partial f}{\partial z} \mathrm{~d} z \\ \\
\Rightarrow & \ \mathrm{d} f (1, 1, 1) = 1 \cdot \mathrm{d} x + (-1) \cdot \mathrm{d} y + 0 \cdot \mathrm{d} z \\ \\
\Rightarrow & \ \textcolor{springgreen}{\boldsymbol{ \mathrm{d} f (1, 1, 1) = \mathrm{d} x – \mathrm{d} y }} \\ \\
\end{aligned}
$$

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