6.2.3 多元函数的微分法则

6.2.3.1 复合函数的微分

1. 一元复合函数

$$\begin{array}{}\text{(6.51a)}& u=f\left(x_1,x_2,\cdots ,x_n\right),x_1=x_1\left(\xi \right),x_2=x_2\left(\xi \right),\cdots ,x_n=x_n\left(\xi \right),\end{array}$$$$\begin{array}{}\text{(6.51b)}& \frac{\partial u}{\partial \xi }=\frac{\partial u}{\partial x_1}\frac{\mathrm{d}{x}_1}{\mathrm{d}\xi }+\frac{\partial u}{\partial x_2}\frac{\mathrm{d}{x}_2}{\mathrm{d}\xi }+\cdots +\frac{\partial u}{\partial x_n}\frac{\mathrm{d}{x}_n}{\mathrm{d}\xi }.\end{array}$$

2. 多元复合函数

$$u=f\left(x_1,x_2,\cdots ,x_n\right),$$$$\begin{array}{}\text{(6.52a)}& x_1=x_1\left(\xi ,\eta ,\cdots ,\tau \right),x_2=x_2\left(\xi ,\eta ,\cdots ,\tau \right),\cdots ,x_n=x_n\left(\xi ,\eta ,\cdots ,\tau \right),\end{array}$$$$\begin{array}{}\text{(6.52b)}& \begin{array}{l}\frac{\partial u}{\partial \xi }=\frac{\partial u}{\partial x_1}\frac{\partial x_1}{\partial \xi }+\frac{\partial u}{\partial x_2}\frac{\partial x_2}{\partial \xi }+\cdots +\frac{\partial u}{\partial x_n}\frac{\partial x_n}{\partial \xi },\\ \frac{\partial u}{\partial \eta }=\frac{\partial u}{\partial x_1}\frac{\partial x_1}{\partial \eta }+\frac{\partial u}{\partial x_2}\frac{\partial x_2}{\partial \eta }+\cdots +\frac{\partial u}{\partial x_n}\frac{\partial x_n}{\partial \eta },\\ \frac{\partial u}{\partial \eta }=\frac{\partial u}{\partial x_1}\frac{\partial x_1}{\partial \eta }+\frac{\partial u}{\partial x_2}\frac{\partial x_2}{\partial \eta }+\cdots +\frac{\partial u}{\partial x_n}\frac{\partial x_n}{\partial \eta }.\end{array}\}\end{array}$$

6.2.3.2 隐函数的微分

(1)若一元函数 $y=f\left(x\right)$ 由方程

$$\begin{array}{}\text{(6.53a)}& F\left(x,y\right)=0\end{array}$$

给出, 则利用 (6.51b) 可对 (6.53a) 进行微分

$$\begin{array}{}\text{(6.53b)}& F_x+F_y{y}^{\prime }=0,\end{array}$$

故

$$\begin{array}{}\text{(6.53c)}& y^{\prime }=-\frac{F_x}{F_y}\left(F_y\ne 0\right).\end{array}$$

用同样的方法可得(6.53b)的微分

$$\begin{array}{}\text{(6.53d)}& F_{xx}+2F_{xy}{y}^{\prime }+F_{yy}{\left(y^{\prime }\right)}^2+F_y{y}^{\prime \prime }=0.\end{array}$$

由(6.53b),有

$$\begin{array}{}\text{(6.53e)}& y^{\prime \prime }=\frac{2F_x{F}_y{F}_{xy}-{\left(F_y\right)}^2{F}_{xx}-{\left(F_x\right)}^2{F}_{yy}}{{\left(F_y\right)}^3}.\end{array}$$

利用类似的方法可计算三阶导数

$$\begin{array}{}\text{(6.53f)}& F_{xxx}+3F_{xxy}{y}^{\prime }+3F_{xyy}{\left(y^{\prime }\right)}^2+F_{yyy}{\left(y^{\prime }\right)}^3+3F_{xy}{y}^{\prime \prime }+3F_{yy}{y}^{\prime }{y}^{\prime \prime }+F_y{y}^{\prime \prime \prime }=0\end{array}$$

由此可把 $y^{\prime \prime \prime }$ 表示出来.

(2) 若多元函数 $u=f\left(x_1,x_2,\cdots ,x_i,\cdots ,x_n\right)$ 由方程

$$\begin{array}{}\text{(6.54a)}& F\left(x_1,x_2,\cdots ,x_i,\cdots ,x_n,u\right)=0\end{array}$$

给出, 可用类似前面的方法来计算偏导数

$$\begin{array}{}\text{(6.54b)}& \frac{\partial u}{\partial x_1}=-\frac{F_{x_1}}{F_u},\frac{\partial u}{\partial x_2}=-\frac{F_{x_2}}{F_u},\cdots ,\frac{\partial u}{\partial x_n}=-\frac{F_{x_n}}{F_u},\end{array}$$

但是此处将会用到 (6.52b). 按同样的方法可计算高阶导数.

(3) 若两个一元函数 $y=f\left(x\right),z=\phi \left(x\right)$ 由

$$\begin{array}{}\text{(6.55a)}& F\left(x,y,z\right)=0\text{ 和 }\mathrm{\Phi }\left(x,y,z\right)=0\end{array}$$

构成的方程组给出. 根据 (6.51b), (6.55a) 的微分为

$$\begin{array}{}\text{(6.55b)}& F_x+F_y{y}^{\prime }+F_z{z}^{\prime }=0,{\mathrm{\Phi }}_x+{\mathrm{\Phi }}_y{y}^{\prime }+{\mathrm{\Phi }}_z{z}^{\prime }=0,\end{array}$$

故

$$\begin{array}{}\text{(6.55c)}& y^{\prime }=\frac{F_z{\mathrm{\Phi }}_x-{\mathrm{\Phi }}_z{F}_x}{F_y{\mathrm{\Phi }}_z-F_z{\mathrm{\Phi }}_y},z^{\prime }=\frac{F_x{\mathrm{\Phi }}_y-F_y{\mathrm{\Phi }}_x}{F_y{\mathrm{\Phi }}_z-F_z{\mathrm{\Phi }}_y}.\end{array}$$

由 $y^{\prime },z^{\prime }$ ,利用 (6.55b) 的微分,用同样的方法可计算二阶导数 $y^{\prime \prime }$ 和 $z^{\prime \prime }$ .

(4) $n$ 个一元函数 令 $y_1=f\left(x\right),y_2=\phi \left(x\right),\cdots ,y_n=\psi \left(x\right)$ 由 $n$ 个方程

$$F\left(x,y_1,y_2,\cdots ,y_n\right)=0,\mathrm{\Phi }\left(x,y_1,y_2,\cdots ,y_n\right)=0,\cdots ,\mathrm{\Psi }\left(x,y_1,y_2,\cdots ,y_n\right)=0$$

(6.56a)

构成的方程组给出. 由 (6.51b), 可得 (6.56a) 的微分

$$\begin{array}{}\text{(6.56b)}& \begin{array}{c}{F}_x+F_{y_1}{y}_1^{\prime }+F_{y_2}{y}_2^{\prime }+\cdots +F_{y_n}{y}_n^{\prime }=0,\\ {\mathrm{\Phi }}_x+{\mathrm{\Phi }}_{y_1}{y}_1^{\prime }+{\mathrm{\Phi }}_{y_2}{y}_2^{\prime }+\cdots +{\mathrm{\Phi }}_{y_n}{y}_n^{\prime }=0,\\ ⋮+⋮+⋮+⋮+⋮+⋮=0,\\ {\mathrm{\Psi }}_x+{\mathrm{\Psi }}_{y_1}{y}_1^{\prime }+{\mathrm{\Psi }}_{y_2}{y}_2^{\prime }+\cdots +{\mathrm{\Psi }}_{y_n}{y}_n^{\prime }=0.\end{array}\}\end{array}$$

解 (6.56b),可得要求的导数 $y_1^{\prime },y_2^{\prime },\cdots ,y_n^{\prime }$ . 用类似的方法可计算高阶导数.

(5) 若两个二元函数 $u=f\left(x,y\right),v=\phi \left(x,y\right)$ 由

$$\begin{array}{}\text{(6.57a)}& F\left(x,y,u,v\right)=0\text{ 和 }\mathrm{\Phi }\left(x,y,u,v\right)=0\end{array}$$

构成的方程组给出. 利用 (6.52b),(6.57a) 的关于 $x,y$ 的微分为

$$\begin{array}{}\text{(6.57b)}& \begin{array}{l}\frac{\partial F}{\partial x}+\frac{\partial F}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial F}{\partial v}\frac{\partial v}{\partial x}=0,\\ \frac{\partial \mathrm{\Phi }}{\partial x}+\frac{\partial \mathrm{\Phi }}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial \mathrm{\Phi }}{\partial v}\frac{\partial v}{\partial x}=0.\end{array}\}\end{array}$$$$\begin{array}{}\text{(6.57c)}& \begin{array}{l}\frac{\partial F}{\partial y}+\frac{\partial F}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial F}{\partial v}\frac{\partial v}{\partial y}=0,\\ \frac{\partial \mathrm{\Phi }}{\partial y}+\frac{\partial \mathrm{\Phi }}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial \mathrm{\Phi }}{\partial v}\frac{\partial v}{\partial y}=0.\end{array}\}\end{array}$$

分别从方程组 (6.57b) 和 (6.57c) 中解出 $\frac{\partial u}{\partial x},\frac{\partial v}{\partial x}$ 和 $\frac{\partial u}{\partial y},\frac{\partial v}{\partial y}$ ,即得一阶偏导数. 利用同样的方法可计算高阶偏导数.

(6) 由 $n$ 个方程构成的方程组给出的 $n$ 个 $m$ 元函数 利用与前面类似的方法可计算一阶和高阶偏导数.