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高数笔记:多元复合函数的隐函数问题

基础知识

隐函数存在定理:$F(x,y)$ 有连续一阶偏导数,且 $F'_y \neq 0$ ,则方程 $F(x,y) = 0$ 确定 $y = y(x)$ ,且 $\frac{dy}{dx} = - \frac{F'_x}{F'_y}$

题型

【例子】 验证 $x^y + y^2 - 1 = 0$$(0,1)$ 附近能确定函数 $ y=y(x) $,求 $\left.\frac{d y}{d x}\right|_{x=0}$$\left.\frac{d^{2} y}{d x^{2}}\right|_{x=0}$

【分析与解答】

$F = x^2 + y^2 -1 = 0$

$F'_y|_{(0,1)} = 2y = 2 \neq 0$ ,所以 $ y $$ x $ 的函数,即 $ y = y(x)$

$$
\frac{d y}{d x}=-\frac{F_{x}^{\prime}}{F_{y}^{\prime}}=-\frac{2 x}{2 y}=-\frac{x}{y}
$$

带入 $(0,1)$ 等于 $0$

$$
\frac{d^{2} y}{d x^{2}}=-\frac{y-x \cdot \frac{d y}{d x}}{y^{2}}
$$

带入 $(0,1)$ 等于 $-1$


【例子】$=x^{2}+y^{2}+z^{2}-4 z=0$$\frac{\partial^{2} z}{\partial x^{2}}$

【分析与解答】

要求 $\frac{\partial^{2} z}{\partial x^{2}}$ 就说明要求 $z = z(x,y)$ 的隐函数。

$$
\begin{align}
&x^2 + y^2 + z^2 - 4z = 0\\
\text{一阶导数}\to & 2x + 0 + 2z \frac{\partial {z} }{\partial x} - 4 \frac{\partial z}{\partial x} = 0\\
\text{二阶导数} \to & 2+2 \frac{\partial z}{\partial x} \cdot \frac{\partial z}{\partial x}+2 z \cdot \frac{\partial^{2} z}{\partial x^{2}}-4 \frac{\partial^{2} z}{\partial x^{2}}=0
\end{align}
$$

所以 $\frac{\partial^2 z}{\partial x ^2} = \frac{(2-z)^{2}+x^{2}}{(2-z)^{3}} $


【例子】$\left\{\begin{array}{l}x u-y v=0 \\y u+x v=1\end{array}\right. \tag{1}$$\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial v}{\partial x} $$ \frac{\partial v}{\partial y}$

由所求应该让 $x,y$ 为自变量,即 $\left\{\begin{array}{l}u=u(x, y) \\v=v(x, y)\end{array}\right.$

$1$ 式对 $x$ 求导,得到:

$$
\left\{\begin{array}{l}
u+x \cdot \frac{\partial u}{\partial x}-y \cdot \frac{\partial V}{\partial x}=0 \\
y \cdot \frac{\partial u}{\partial x}+v+x \frac{\partial v}{\partial x}=0
\end{array}\right.
$$

解得

$$
\left\{\begin{array}{l}
\frac{\partial u}{\partial x}=-\frac{x u+y v}{x^{2}+y^{2}} \\
\frac{\partial x}{\partial x}=\frac{y u-x x}{x^{2}+y^{2}}
\end{array}\right.
$$

方程组求导,用直接法,然后解方程。

【例子】 证明 $\varphi(c x-a z, c y-b z)=0$ 所确定的函数 $z = f(x,y)$ 满足 $a \frac{\partial z}{\partial x}+b \frac{\partial z}{\partial y}=c$ 。已知 $\varphi (u,v)$ 具有连续偏导数。

【分析与解答】

$\varphi(c x-a z, c y-b z)=0$ 两端对 $x$ 求导。得到:

$$
\begin{align}
&\varphi _1' (cx - az)' + \varphi _2'(y - bz)' = 0\\
\to & \varphi _1' (c - a \frac{\partial z}{\partial x} ) + \varphi _2'(-b \frac{\partial z}{\partial x} ) = 0
\end{align}
$$

由此可得:

$$
c \varphi_{1}^{\prime}=\left(a_{1} \varphi_{1}^{\prime}+b \varphi_{2}^{\prime}\right) \cdot \frac{\partial z}{\partial x}\\
{\color{red}ac \varphi_{1}^{\prime}}=\left(a_{1} \varphi_{1}^{\prime}+b \varphi_{2}^{\prime}\right) \cdot a\frac{\partial z}{\partial x}
$$

所以

$$
a \frac{\partial z}{\partial x}+b \frac{\partial z}{\partial y}=\frac{{\color{red}ac \varphi_{1}^{\prime}}}{a \varphi_{1}^{\prime}+b \varphi_{2}^{\prime}} + \frac{bc \varphi _2'}{a \varphi_{1}^{\prime}+b \varphi_{2}^{\prime}} = c
$$

【例子】 已知 $y = f(x,t)$$t = t(x,y)$$F(x,y,t) = 0$ 确定的函数,$f,F$ 都有一阶连续偏导数,求 $\frac{dy}{dx} $

根据题目可猜测 $y$$ x $ 的一元函数。

$$
\begin{align}
\frac{\partial y}{\partial x} &= f_1' + f_2' \frac{\partial t}{\partial x} \tag{1}
\end{align}
$$

根据 $F(x,y,t) = 0$ 两端求导可得:

$$
F_1' \cdot 1 + F_2' \cdot \frac{\partial y}{\partial x} + F_3' \frac{\partial t}{\partial x} = 0 \tag{2}
$$

联立 $ 1,2 $ 可得

$$
\begin{align}
\frac{\partial y}{\partial x} &= f_1' - f_2' \cdot \frac{F_1' + F_2' \frac{\partial y}{\partial x} }{F_3'}\\
\text{两边乘以} F_3' \to F_3' \frac{\partial y}{\partial x} &= F_3'f_1' - f_2' F_{1}^{\prime}- f_2' F_{2}^{\prime} \frac{\partial y}{\partial x} = 0
\end{align}
$$

提出 $\frac{dy}{dx} $ 可得

$$
\frac{dy}{dx} = \frac{F_3'f_1' - f_2' F_1'}{F_3' +f_2'F_2'}
$$

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