微甲II-隐函数求导

隐函数的偏导数

定理一

设$F(x,y)$在$P(x_0,y_0)$的某一邻域内满足以下条件

$$\begin{aligned} \begin{cases} \text{具有连续的偏导数}\\ F(x_0,y_0)=0\\ F_y(x_0,y_0)\ne 0 \end{cases} \end{aligned}$$

则$F(x,y)=0$在点$x_0$的某一邻域内可以确定$y=f(x)$满足$y_0=f(x_0)$，并且有连续导数

$$\frac{dy}{dx}=-\frac{F_x}{F_y}\text{（隐函数求导公式）}$$

证明略。

推导

设$F(x,f(x))=0$，两边对$x$求导，有$\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{F_x}{F_y}$​。

若$F(x,y)$二阶偏导数也连续，则可以求隐函数的二阶导数

$$\begin{aligned} \frac{\mathrm{d}^2y}{\mathrm{d}x^2}& =\frac\partial{\partial x}(-\frac{F_x}{F_y})+\frac\partial{\partial y}(-\frac{F_x}{F_y})\cdot\frac{\mathrm{d}y}{\mathrm{d}x} \\ &=-\frac{F_{xx}F_y-F_{yx}F_x}{F_y^2}-\frac{F_{xy}F_y-F_{yy}F_x}{F_y^2}(-\frac{F_x}{F_y}) \\ &=-\frac{F_{xx}F_y^2-2F_{xy}F_xF_y+F_{yy}F_x^2}{F_y^3} \end{aligned}$$

定理二

若函数$F(x,y,z)$满足：

$$\begin{aligned} \begin{cases} \text{在点}P(x_0,y_0,z_0)\text{的某邻域内具有连续偏导数}\\ F(x_0,y_0,z_0)=0\\ F_z(x_0,y_0,z_0)\ne 0 \end{cases} \end{aligned}$$

则相似的有

$$\frac{\partial z}{\partial x}=-\frac{F_x}{F_z},\frac{\partial z}{\partial y}=-\frac{F_y}{F_z}$$

证明略

推导

设$z=f(x,y)$是方程$F(x,y,z)=0$确定隐函数，则

$$F(x,y,f(x,y))\equiv 0\\[2pt] \quad\downarrow \text{两边对x求偏导}\\[2pt] F_x+F_z\frac{\partial z}{\partial x}=0\\[2pt] \ \quad\downarrow \text{在邻域内}F_z\ne0\\[2pt] \frac{\partial z}{\partial x}=-\frac{F_x}{F_z}$$

方程组所确定的隐函数组及其导数

$$\begin{cases} F(x,y,u,v)=0\\ G(x,y,u,v)=0 \end{cases} \Rightarrow \begin{cases} u=u(x,y)\\ v=v(x,y) \end{cases}$$

由$F,G$的偏导数组成的行列式

$$J=\frac{\partial(F,G)}{\partial(u,v)}= \begin{vmatrix} F_u & F_v\\ G_u & G_v \end{vmatrix}$$

称为$F$、$G$的雅可比行列式

定理三

与定理一、定理二相似的，将第三条条件换为

$$J|_P=\frac{\partial(F,G)}{\partial(u,v)}\bigg|_P\ne0$$

有偏导数公式：

$$\frac{\partial u}{\partial x}=-\frac1J\frac{\partial(F,G)}{\partial(x,v)}=-\frac1{\begin{vmatrix}F_u&F_v\\G_u&G_v\end{vmatrix}}\begin{vmatrix}F_x&F_v\\G_x&G_v\end{vmatrix}$$

$$\frac{\partial u}{\partial y}=-\frac1J\frac{\partial(F,G)}{\partial(y,v)}=-\frac1{\begin{vmatrix}F_u&F_v\\G_u&G_v\end{vmatrix}}\begin{vmatrix}F_y&F_v\\G_y&G_v\end{vmatrix}$$

$$\begin{aligned}&\frac{\partial v}{\partial x}=-\frac1J\frac{\partial(F,G)}{\partial(u,x)}=-\frac1{\begin{vmatrix}F_u&F_v\\G_u&G_v\end{vmatrix}}\begin{vmatrix}F_u&F_x\\G_u&G_x\end{vmatrix}\end{aligned}$$

$$\begin{aligned}&\frac{\partial v}{\partial y}=-\frac1J\frac{\partial(F,G)}{\partial(u,y)}=-\frac1{\begin{vmatrix}F_u&F_v\\G_u&G_v\end{vmatrix}}\begin{vmatrix}F_u&F_y\\G_u&G_y\end{vmatrix}\end{aligned}$$

例

设$xu-yv=0,yu+xv=1$，求$\frac{\partial u}{\partial x},\frac{\partial v}{\partial x}.$

$$\begin{cases} x\displaystyle\frac{\partial u}{\partial x}-y\frac{\partial v}{\partial x}=-u\\[2pt] t\displaystyle\frac{\partial u}{\partial x}+x\frac{\partial v}{\partial x}=-v \end{cases}$$

由题设知$J=\begin{vmatrix}x&-y\\-y&x\end{vmatrix}=x^2+y^2\ne0$​

$$\text{故有}\left\{\begin{array}{c}\displaystyle\frac{\partial u}{\partial x}=\frac{1}{J} \begin{vmatrix}-u&-y\\-v&x\end{vmatrix}=-\frac{xu+yv}{x^2+y^2}\\[2pt] \displaystyle\frac{\partial v}{\partial x}=\frac{1}{J} \begin{vmatrix}x&-u\\y&-v\end{vmatrix}=- \frac{xv-yu}{x^2+y^2}\end{array}\right.$$