20211012隐函数求导公式

隐函数求导公式

一、一个方程的情形

隐函数存在定理1：

设函数 \(\displaystyle F(x, y)\) 在点 \(\displaystyle P(x_0, y_0)\) 的某一邻域内具有连续的偏导数，且 \(\displaystyle F(x_0, y_0) = 0, F_y(x_0, y_0) \neq 0\) ，则方程 \(\displaystyle F(x, y) = 0\) 在点 \(\displaystyle (x_0, y_0)\) 的某一邻域内恒能确定一个连续且具有连续导数的函数 \(\displaystyle y=f(x)\) ，它满足条件 \(\displaystyle y_0 = f(x_0)\) ，并有

\[\cfrac{\mathrm{d}y}{\mathrm{d}x} = - \cfrac{F_x}{F_y} \tag{1} \label{eq1} \]

公式 \(\displaystyle \eqref{eq1}\) 就是隐函数求导公式。

将方程 \(\displaystyle F(x, y) = 0\) 所确定的函数 \(\displaystyle y = f(x)\) 代入 \(\displaystyle F(x, y) = 0\) ，的恒等式：

\[F(x, f(x)) \equiv 0 \nonumber \]

其左端可看作是 \(\displaystyle x\) 的一个复合函数，求这个函数的全导数，由于恒等式两端求导后仍然恒等，即得：

\[\cfrac{\partial F}{\partial x} + \cfrac{\partial F}{\partial y} \cfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \nonumber \]

因为 \(\displaystyle F_y\) 连续，且 \(\displaystyle F_y (x_0, y_0) \neq 0\) ，所以存在 \(\displaystyle (x_0, y_0)\) 的一个邻域，在这个邻域内 \(\displaystyle F_y \neq 0\) ，于是得

\[\cfrac{\mathrm{d}y}{\mathrm{d}x} = - \cfrac{F_x}{F_y} \nonumber . \]

如果 \(\displaystyle F(x,y)\) 的二阶偏导数也都连续，可以把等式 \(\displaystyle \eqref{eq1}\) 的两端看做 \(\displaystyle x\) 的复合函数而再一次求导：

\[\begin{align} \cfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} &= \cfrac{\partial}{\partial x} \left( -\cfrac{F_x}{F_y} \right) + \cfrac{\partial}{\partial x} \left( -\cfrac{F_x}{F_y} \right) \cfrac{\mathrm{d}y}{\mathrm{d}x} \nonumber \\ &= - \cfrac{F_{xx}F_y - F_{yx}F_x}{F^2_y} - \cfrac{F_{xy}F_y - F_{yy}F_x}{F^2_y} \left(-\cfrac{F_x}{F_y} \right) \nonumber \\ &= - \cfrac{F_{xx}F_y - 2F_{xy} F_x F_y + F_{yy}F^2_x}{F^3_y} \nonumber \end{align} \]

隐函数存在定理2：

设函数 \(\displaystyle F(x, y, z)\) 在点 \(\displaystyle P(x_0, y_0, z_0)\) 的某一邻域内具有连续的偏导数，且 \(\displaystyle F(x_0, y_0, z_0) = 0, \, F_z(x_0, y_0, z_0) \neq 0\) ，则方程 \(\displaystyle F(x, y, z) = 0\) 在点 \(\displaystyle (x_0, y_0, z_0)\) 的某一邻域内恒能唯一确定一个连续且具有连续导数的函数 \(\displaystyle z=f(x, y)\) ，它满足条件 \(\displaystyle z_0 = f(x_0, y_0)\) ，并有

\[\cfrac{\partial z}{\partial x} = - \cfrac{F_x}{F_z} ,\, \cfrac{\partial z}{\partial y} = - \cfrac{F_y}{F_z} . \tag{2} \label{eq2} \]

例题：设 \(\displaystyle x^2 + y^2 + z^2 - 4z = 0\) ，求 \(\displaystyle \cfrac{\partial^2 z}{\partial x^2}\) .

解： 设 \(\displaystyle F(x, y, z) = x^2 + y^2 + z^2 - 4z\) ，则 \(\displaystyle F_x = 2x, \, F_z = 2z - 4\) .当 \(\displaystyle z \neq 2\) 时，应用公式 \(\eqref{eq2}\) 得，

\[\cfrac{\partial z}{\partial x} = \cfrac{x}{z - 2} \nonumber \]

再一次对 \(x\) 求偏导数，得：

\[\cfrac{\partial^2 z}{\partial x^2} = \cfrac{(2 - z) + x \cfrac{\partial z}{\partial x}}{(2 - z)^2} = \cfrac{(2 - z) + x \left(\cfrac{x}{2 - z} \right)}{(2 - z)^2} = \cfrac{(2 - z)^2 + x^2}{(2 - z)^3} \nonumber \]

二、方程组的情形

隐函数存在定理3：

设函数 \(\displaystyle F(x, y, u, v)、G(x, y, u, v)\) 在点 \(\displaystyle P(x_0, y_0, u_0, v_0)\) 的某一邻域内具有对各个变量的连续偏导数，又 \(\displaystyle F(x_0, y_0, u_0, v_0) = 0, \, G(x_0, y_0, u_0, v_0) = 0\) ，且偏导数所组成的函数行列式（或称为雅克比 \(Jacobi\) 式）

\[J = \cfrac{\partial (F, G)}{\partial (u, v)} = \left| \begin{array}{cc} \cfrac{\partial F}{\partial u} & \cfrac{\partial F}{\partial v} \\ \cfrac{\partial G}{\partial u} & \cfrac{\partial G}{\partial v} \end{array} \right| \nonumber \]

在点 \(\displaystyle P(x_0, y_0, u_0, v_0)\) 不等于零，则方程组 \(\displaystyle F(x, y, u, v)=0,G(x, y, u, v)=0\) 在点 \((x_0, y_0, u_0, v_0)\) 的某一邻域内恒能唯一确定一组连续且具有连续偏导数的函数 \(\displaystyle u = u(x, y), v = v(x, y)\) ，它们满足条件 \(\displaystyle u_0 = u(x_0, y_0), v_0 = v(x_0, y_0)\) ，并有

\[\begin{align} \cfrac{\partial u}{\partial x} &= -\cfrac{1}{J} \cfrac{\partial (F, G)}{\partial (x, v)} = -\cfrac{\left| \begin{array}{cc} F_x & F_v \\ G_x & G_v\end{array} \right|}{\left| \begin{array}{cc} F_u & F_v \\ G_u & G_v\end{array} \right|}, \nonumber \\ \cfrac{\partial v}{\partial x} &= -\cfrac{1}{J} \cfrac{\partial (F, G)}{\partial (u, x)} = -\cfrac{\left| \begin{array}{cc} F_u & F_x \\ G_u & G_x\end{array} \right|}{\left| \begin{array}{cc} F_u & F_v \\ G_u & G_v\end{array} \right|}, \nonumber \\ \cfrac{\partial u}{\partial y} &= -\cfrac{1}{J} \cfrac{\partial (F, G)}{\partial (y, v)} = -\cfrac{\left| \begin{array}{cc} F_y & F_v \\ G_y & G_v\end{array} \right|}{\left| \begin{array}{cc} F_u & F_v \\ G_u & G_v\end{array} \right|}, \nonumber \\ \cfrac{\partial v}{\partial y} &= -\cfrac{1}{J} \cfrac{\partial (F, G)}{\partial (u, y)} = -\cfrac{\left| \begin{array}{cc} F_u & F_y \\ G_u & G_y\end{array} \right|}{\left| \begin{array}{cc} F_u & F_v \\ G_u & G_v\end{array} \right|}. \nonumber \end{align} \tag{3} \label{eq3} \]