3-21 微积分笔记

一个方程确定的隐函数

【定理1：隐函数存在定理】设$F(x,y,z)$在$P_0(x_0,y_0,z_0)$附近满足
- $F(x,y,z)$具有一阶连续偏导数
- $F(x_0,y_0,z_0)=0$
- $F'_z(x_0,y_0,z_0)=F'_z|_{P_0}\neq 0$
则在$(x_0,y_0)$的某邻域上存在唯一的$z=f(x,y)$，且$z_0=f(x_0,y_0)$，且$F(x,y,f(x,y))=0$，且$z=f(x,y)$具有连续偏导数，且

$$\frac{\partial z}{\partial x}=-\frac{F'_x(x,y,z)}{F'_z(x,y,z)},\qquad \frac{\partial z}{\partial y}=-\frac{F'_y(x,y,z)}{F'_z(x,y,z)}$$
这是因为：
$$F(x,y,f(x,y))\equiv 0$$
两边对$x$求偏导
$$F'_x(x,y,z)+F'_z(x,y,z)\cdot\frac{\partial z}{\partial x}=0$$
即为我们想要验证的式子。

例1、验证：$f(x,y)=xy+2^x-2^y$在$(0,0)$处的某邻域确定唯一隐函数$y=y(x)$；并求导。

解：易知$F(x,y)=xy+2^x-2^y$在$R^2$有一阶连续偏导，且$F(0,0)=0,F'_y|_{(0,0)}\neq 0$，故由隐函数存在定理，在$x_0$附近存在唯一、单值的隐函数$y=y(x)$，且$y$可导。

$$\frac{dy}{dx}=-\frac{F'_x}{F'_y}=-\frac{y+2^x\ln 2}{x-2^y\ln 2}$$

例2、设$F(\frac xz,\frac zy)=0$确定隐函数$z=z(x,y)$，且$F$具有连续偏导数，求$z'_x,z'_y$

解：由$F(\frac xz,\frac zy)=0$得

$$\frac{\partial z}{\partial x}=-\frac{F'_x}{F'_z}=-\frac{F_1'\cdot \frac1z + F_2'\cdot 0}{F_1'(-\frac x{z^2})+F_2'\cdot \frac1y}$$

$$\frac{\partial z}{\partial y}=-\frac{F_2'(-\frac z{y^2})}{F'_1(-\frac x {z^2}+F'_2\cdot\frac1y)}$$

例3、设$z=z(x,y)$由方程$x^2+y^2-z=\varphi(x+y+z)$所确定，其中$\varphi$有连续导数。且$\varphi \neq -1$，求$dz$.

解法一：由公式得

$$\frac{\partial z}{\partial x}=-\frac{2x-\varphi'(x+y+z)}{-1-\varphi'(x+y+z)}=\frac{2x-\varphi'(x+y+z)}{1+\varphi'(x+y+z)} \\ \frac{\partial z}{\partial y}=\frac{2y-\varphi'(x+y+z)}{1+\varphi'(x+y+z)}$$
解法二：方程两边对$x$求导
$$2x-z'_x=\varphi'(x+y+z)(1+z'_x)$$
易知 $$z_x'=\frac{2x-\varphi'(x+y+z)}{1+\varphi'(x+y+z)}$$

多个方程确定隐函数组求偏导

【定义：雅可比行列式】
设$F_i(x_1,\cdots,x_n,y_1,\cdots,y_m)$（其中$i=1,\cdots,m$）具有连续偏导数，记

$$J=\frac{\partial (F_1,\cdots, F_n)}{\partial (y_1,\cdots ,y_m)} = \left|\begin{array}{cccc} \frac{\partial F_1}{\partial y_1} & \cdots & \frac{\partial F_m}{\partial y_1} \\ \cdots & \cdots & \cdots \\ \frac{\partial F_1}{\partial y_m} & \cdots & \frac{\partial F_m}{\partial y_m} \end{array}\right|$$

称为雅可比行列式。

【定理2】设$F(x,y,u,v),G(x,y,u,v)$在$P_0(x_0,y_0,u_0,v_0)$的某邻域满足
- $F,G$具有一阶连续偏导数
- $F(P_0)=0,G(P_0)=0$
- $J=\frac{\partial (F,G)}{\partial (u,v)} \Big|_{P_0} \neq 0$
则在$(x_0,y_0)$附近，唯一存在$u=u(x,y),v=v(x,y)$，且$u_0=u(x_0,y_0),v_0=v(x_0,y_0)$，且$F(x,y,u(x,y),v(x,y))\equiv 0,G(x,y,u(x,y),v(x,y))\equiv 0$，且$u(x,y),v(x,y)$具有连续偏导数。

$$\frac{\partial u}{\partial x}=-\frac{\partial (F,G)}{\partial (x,v)}/\frac{\partial (F,G)}{\partial (u,v)}\\ \frac{\partial u}{\partial y}=-\frac{\partial (F,G)}{\partial (y,v)}/\frac{\partial (F,G)}{\partial (u,v)}\\ \frac{\partial v}{\partial x}=-\frac{\partial (F,G)}{\partial (u,x)}/\frac{\partial (F,G)}{\partial (u,v)}\\ \frac{\partial v}{\partial y}=-\frac{\partial (F,G)}{\partial (u,y)}/\frac{\partial (F,G)}{\partial (u,v)}$$

$$\frac{\partial (F,G)}{\partial (u,v)}=\left|\begin{array}{cccc} F'_u & F'_v \\ G'_u & G'_v \end{array}\right| = \left|\begin{array}{cccc} F'_u & G'_u \\ F'_v & G'_v \end{array}\right|$$

$$\frac{\partial(F,G)}{\partial(x,u)}=\left|\begin{array}{cccc} F'_x & F'_v \\ G'_x & G'_v \end{array}\right|$$