## 题型

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

【分析与解答】

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

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

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

【分析与解答】

\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}

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

$1$ 式对 $x$ 求导，得到：

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

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

【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}

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

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

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

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

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

\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}

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