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：

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

接着，我们对题目所给函数两端做变形处理：

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

又因为：

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

于是：

$$\begin{array}{rl}& {[\ln f]}_x^{\prime }={[\frac{1}{z}\ln x–\frac{1}{z}\ln y]}_x^{\prime }\\ \\ \Rightarrow & \frac{f_x^{\prime }}{f}=\frac{1}{x}\\ \\ \Rightarrow & x=1\\ \\ \Rightarrow & f_x^{\prime }=\frac{\partial f}{\partial x}=1\end{array}$$

于是：

$$\begin{array}{rl}& {[\ln f]}_y^{\prime }={[\frac{1}{z}\ln x–\frac{1}{z}\ln y]}_y^{\prime }\\ \\ \Rightarrow & \frac{f_y^{\prime }}{f}=\frac{1}{y}\\ \\ \Rightarrow & y=1\\ \\ \Rightarrow & f_y^{\prime }=\frac{\partial f}{\partial y}=-1\end{array}$$

于是：

$$\begin{array}{rl}& {[\ln f]}_z^{\prime }={[\frac{1}{z}\ln x–\frac{1}{z}\ln y]}_z^{\prime }\\ \\ \Rightarrow & \{\begin{array}{l}x=1\\ y=1\end{array}\\ \\ \Rightarrow & {[\ln f]}_z^{\prime }={[\frac{1}{z}\ln 1–\frac{1}{z}\ln 1]}_z^{\prime }\\ \\ \Rightarrow & \frac{f_z^{\prime }}{f}={\left[0\right]}_z^{\prime }\\ \\ \Rightarrow & f_z^{\prime }=\frac{\partial f}{\partial z}=0\end{array}$$

综上可知：

$$\begin{array}{rl}& \mathrm{d}f=\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y+\frac{\partial f}{\partial z}\mathrm{d}z\\ \\ \Rightarrow & \mathrm{d}f(1,1,1)=1\cdot \mathrm{d}x+(-1)\cdot \mathrm{d}y+0\cdot \mathrm{d}z\\ \\ \Rightarrow & \mathbf{d}\mathit{f}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{d}\mathit{x}–\mathbf{d}\mathit{y}\\ \end{array}$$

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