*全微分定义

$z=f(x,y)在(x,y)处的全增量\Delta a=A\Delta x+B\Delta y+o(\rho )。A,B不依赖于\Delta x、\Delta y且仅与x,y有关。则称z=f(x,y)在(x,y)处可微。$

$称A\Delta x+B\Delta y为z=f(x,y)在(x,y)处的全微分。$

$\begin{array}{rl}记作dz& =A\Delta x+B\Delta y\\ & =\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy\end{array}$

f(x,y)可微性判断

1. 写出全增量$\Delta z=f(x_0+\Delta x,y_0+\Delta y)-f(x_0,y_0);$
2. 写出线性增量$A\Delta x+B\Delta y，其中A=f_x^{\prime }(x_0,y_0)，B=f_y^{\prime }(x_0,y_0);$
3. 作极限$\underset{\genfrac{}{}{0ex}{}{\Delta x\to 0}{\Delta y\to 0}}{\lim }\frac{\Delta z-(A\Delta x+B\Delta y)}{\sqrt{(\Delta x{)}^2+(\Delta y{)}^2}}=0，则z=f(x,y)在(x_0,y_0)处可微。$

隐函数存在定理

$$\begin{gathered} 设函数F(x,y)在点P(x_0,y_0)的某一邻域内具有连续偏导数, \\ F(x_0,y_0)=0且F_y^{\prime }(x_0,y_0)\ne 0,则方程F(x,y)=0在点(x_0,y_0)的某一邻域内 \\ 能唯一确定一个连续且具有连续导数的函数y=f(x), \\ 它满足条件y_0=f(x_0),并有\frac{dy}{dx}=-\frac{F_x^{\prime }}{F_y^{\prime }}. \end{gathered}$$

这里
$F(x_0,y_0)=0$：函数值存在；
$F_y^{\prime }(x_0,y_0)\ne 0$：偏导数值存在；

偏导数

f具有二阶连续偏导数，求导次序可以交换。

定义法

$z=f(x,y)在(x_0,y_0)处对x的偏导数;$

$f_x^{\prime }(x_0,y_0)=\underset{\Delta x\to 0}{\lim }\frac{f(x_0+\Delta x,y_0)-f(x_0,y_0)}{\Delta x}$

公式法

$f_x(x,y)直接对x求导$

偏导数连续性判断

1. 定义法求$f_x^{\prime }(x_0,y_0),f_y^{\prime }(x_0,y_0);$
2. 公式法求$f_x^{\prime }(x,y),f_y^{\prime }(x,y);$
3. 计算$\underset{\genfrac{}{}{0ex}{}{\Delta x\to x_0}{\Delta y\to y_0}}{\lim }{f}_x^{\prime }(x,y),\underset{\genfrac{}{}{0ex}{}{\Delta x\to x_0}{\Delta y\to y_0}}{\lim }{f}_y^{\prime }(x,y);$
4. 若$\underset{\genfrac{}{}{0ex}{}{\Delta x\to x_0}{\Delta y\to y_0}}{\lim }{f}_x^{\prime }(x,y)=f_x^{\prime }(x_0,y_0),\underset{\genfrac{}{}{0ex}{}{\Delta x\to x_0}{\Delta y\to y_0}}{\lim }{f}_y^{\prime }(x,y)=f_y^{\prime }(x_0,y_0)，则z=f(x,y)在(x_0,y_0)处连续。$

极值

无条件极值

1. 二元函数取极值的必要条件；
   $设z=f(x,y)在(x_0,y_0)\{\begin{array}{l}一阶偏导数存在\\ 取极值\end{array}，则f_x^{\prime }(x_0,y_0)=0,f_y^{\prime }(x_0,y_0)=0.$
2. 二元函数取极值的充分条件；
   $\{\begin{array}{l}{f}_{xx}^{\prime \prime }(x_0,y_0)=a\\ f_{xy}^{\prime \prime }(x_0,y_0)=b\\ f_{yy}^{\prime \prime }(x_0,y_0)=c\end{array},\Delta =ac-b^2\{\begin{array}{l}>0\Rightarrow 极值\{\begin{array}{l}a<0\Rightarrow 极大值\\ a>0\Rightarrow 极小值\end{array}\\ <0\Rightarrow 非极值\\ =0\Rightarrow 方法失效\end{array}$

条件极值与拉格朗日乘数法

$求u=f(x,y,z)在条件\{\begin{array}{l}\phi (x,y,z)=0\\ \Phi (x,y,z)=0\end{array}下的最值。$

1. $构造辅助函数F(x,y,z,\lambda ,\mu )=f(x,y,z)+\lambda \phi (x,y,z)+\mu \Phi (x,y,z).$
2. 令$\{\begin{array}{l}{F}_x^{\prime }=f_x^{\prime }+\lambda {\Phi }_x^{\prime }+\mu {\phi }_x^{\prime }=0\\ F_y^{\prime }=f_y^{\prime }+\lambda {\Phi }_y^{\prime }+\mu {\phi }_y^{\prime }=0\\ F_z^{\prime }=f_z^{\prime }+\lambda {\Phi }_z^{\prime }+\mu {\phi }_z^{\prime }=0\\ F_{\lambda }^{\prime }=\Phi (x,y,z)=0\\ F_{\mu }^{\prime }=\phi (x,y,z)=0\end{array}$
3. $解上述方程得若干点(x_0,y_0,z_0)\dots (x_n,y_n,z_n)，取u_{max}or{u}_{min}.$

二元函数在区域D下的最值

①只需求出D的内部及 $\Rightarrow 无条件极值问题，求出f_x^{\prime }=0,f_y^{\prime }=0的所有可疑点。$
②D的边界$\Rightarrow \{\begin{array}{l}直接带入求驻点或导数不存在的点\\ 拉格朗日乘数法\end{array}$
上的驻点和导数不存在的点(不用判断它们是否为极值点),
并求出这些点的函数值,然后比较它们的大小,求出最大值和最小值.