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高数笔记：多元复合函数求导问题

基础知识

$$\begin{array}{l} \frac{\partial z}{\partial x}=\frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x}+\frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x}=f_{1}^{\prime} \cdot \frac{\partial u}{\partial x}+f_{2}^{\prime} \cdot \frac{\partial v}{\partial x} \\ \frac{\partial z}{\partial y}=\frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial y}+\frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial y}=f_{1}^{\prime} \cdot \frac{\partial u}{\partial y}+f_{2}^{\prime} \cdot \frac{\partial v}{\partial y} \end{array}$$

【要点】

1. 层层展开求导。
2. 导数的复合路径图不变。

题型

复合函数偏导数

【例子】$w = f(x+y+z, xyz)$$\frac{\partial w}{\partial x}$$ \frac{\partial ^2w}{\partial x \partial z}$

【分析与解答】

f
x+y+z
x
y
z
xyz
x
y
z
\begin{align} \frac{\partial w}{\partial x} &= f_1' (x+y+z)' + f_2' (xyz)'\\ &= f_1' + f_2' yz \end{align}
\begin{align} \frac{\partial ^2w}{\partial x \partial z} &=(f_{11}^{''}1 + f_{12}^{''} xy) + [(f_{21}^{''} 1 + f_{22}^{11})yz + f_2' y]\\ &= f_11^{''} + (xy+yz)f_{12}^{''} + xy^2z f_{22}^{''}+yf_2' \end{align}

【例子】$u=f(x,y,z), y = g (x,t), t = h (x,z)$ ，都有一阶连续偏导数，求 $\frac{\partial u}{\partial x}, \frac{\partial u}{\partial z}$

【分析与解答】

u = f
x
y = g
x
t = h
x
z
t = h
x
z

$x$ 一共有 4 条路径。一层一层展开

\begin{align} \frac{\partial u}{\partial x} &= f_1' \cdot (x)' + f_2' \cdot (y)' + f_3' \cdot (z)'\\ &= f_1' + f_2'(g_1' + g_2' \cdot t') + 0 \\ &= f_1' + f_2' (g_1' + g_2' \cdot (h_1' )) \\ &= f_1' + f_2'g_1' + f_2'g_2'h_1' \end{align}

【例子】 z = f(u, x, y)$$u = xe^y ，求 \frac{\partial ^2 z}{\partial x \partial y} 【分析与解答】 分析复合路径 z = f u x y x y 先求 \frac{\partial z}{\partial x} z\to y 的分支就不考虑了，一定是 0$$ \begin{align} \frac{\partial z}{\partial x} &= f_1'\cdot u' + f_2'\cdot x'\\ &= f_1'e^y + f_2' \end{align} $$然后求 \frac{\partial ^2 z}{\partial x \partial y}$$ \begin{align} \frac{\partial ^2 z}{\partial x \partial y} &= [(f_{11}^{''}u' + f_{13}^{''}y')e^y + f_1' e^y] + (f_{21}^{''}u'+f_{23}^{''}y')\\ &= [(f_{11}^{''}xe^y + f_{13}^{''}y')e^y + f_1' e^y] + (f_{21}^{''}xe^y+f_{23}^{''}y')\\ &= xe^{2y}f_{11}^{''} + e^y f_{13}^{''} + e^y f_1' + xe^yf_{21}^{''} + f_{23}^{''} \end{align} $$【例子】 用变换 \left\{\begin{array}{l}u=x-2 y \\v=x+a y\end{array}\right. 可把 方程 6 \frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial x \partial y}-\frac{\partial^{2} z}{\partial y^{2}}=0 化简为 \frac{\partial^{2} z}{\partial u \partial v}=0，求 a$$z 有二阶连续偏导数 。

【分析与解答】

6 \frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial x \partial y}-\frac{\partial^{2} z}{\partial y^{2}}=0$$z 关于 x,y 的方程，替换后变成了 z 关于 u,v 的方程。 分析复合路径： • z • u • x • y • v • x • y$$ \begin{align} \frac{\partial z}{\partial x} &= {\color{red}\frac{\partial z}{\partial u} \cdot 1} + {\color{blue}\frac{\partial z}{\partial v} \cdot 1} \end{align}  \begin{align} \frac{\partial ^2 z}{\partial x^2} &= {\color{red}[\frac{\partial z}{\partial u}(\frac{\partial z}{\partial u} \cdot 1 + \frac{\partial z}{\partial v} \cdot 1) ]} + {\color{blue}[\frac{\partial z}{\partial v} (\frac{\partial z}{\partial u} \cdot 1 + \frac{\partial z}{\partial v} \cdot 1)]}\\ &= \frac{\partial ^2z }{\partial u^2} + 2 \frac{\partial ^2 z}{\partial uv} + \frac{\partial ^2 z}{\partial v^2} \tag{1} \end{align} $$PS：二阶偏导连续所以可以合并混合导$$ \begin{align} \frac{\partial z}{\partial y} &= \frac{\partial z}{\partial u} \cdot (-2) + \frac{\partial z}{\partial v} \cdot a \end{align} $$套娃开始（$$ \begin{align} \frac{\partial^2 z}{\partial x \partial y} &= -2 \frac{\partial z}{\partial u} [{\color{red}\frac{\partial z}{\partial u} \cdot 1} + {\color{blue}\frac{\partial z}{\partial v} \cdot 1}] + a \frac{\partial z}{\partial v} [{\color{red}\frac{\partial z}{\partial u} \cdot 1} + {\color{blue}\frac{\partial z}{\partial v} \cdot 1}]\\ &= -2 \frac{\partial ^2 z}{\partial u^2} + (a -2) \frac{\partial ^2 z}{\partial u \partial v} + a \frac{\partial ^2 z}{\partial v^2}\tag{2} \end{align}  \begin{align} \frac{\partial^2 z}{\partial^2 y} &= -2 \frac{\partial z}{\partial u} [\frac{\partial z}{\partial u} \cdot (-2) + \frac{\partial z}{\partial v} \cdot a] + a \frac{\partial z}{\partial v} [\frac{\partial z}{\partial u} \cdot (-2) + \frac{\partial z}{\partial v} \cdot a]\\ &= 4 \frac{\partial ^2 z}{\partial u^2} -4a \frac{\partial ^2 z }{\partial u \partial v} + a^2 \frac{\partial ^2 z}{\partial v ^2} \tag{3} \end{align} $$1,2,3 式代入，得到：$$ (10+5a) \frac{\partial ^2 z}{\partial u \partial v} + (6 + a - a^2) \frac{\partial ^2 z}{\partial v^2} $$想要 \frac{\partial ^2 z}{\partial u \partial v} = 0 ，只需 6 + a - a^2 = 0 \and 10 + 5a \ne 0。方程解为$$ a = 3\$