1、二元函数偏导数定义：设函数z=f(x,y)在点$(x_{0},y_{0})$的某邻域有定义，固定y=$y_{0}$，是x从$x_{0}$变到$x_{0}+\Delta x$时，函数的变化为$f(x_{0}+\Delta x,y_{0})-f(x_{0},y_{0})$。如果极限
\[\lim_{\Delta x \rightarrow 0}\frac{f(x_{0}+\Delta x,y_{0})-f(x_{0},y_{0})}{\Delta x}\]
存在，则称此极限为z=f(x,y)在$(x_{0},y_{0})$对x的偏导数，记做$\frac{\partial z}{\partial x}|_{y=y_{0}}^{^{x=x_{0}}}$
2、设函数z=f(x,y)在点$M_{0}(x_{0},y_{0})$的某邻域内存在二阶偏导数$\frac{\partial z}{\partial x\partial y}$和$\frac{\partial z}{\partial y\partial x}$。如果$\frac{\partial z}{\partial x\partial y}$和$\frac{\partial z}{\partial y\partial x}$都在$M_{0}(x_{0},y_{0})$点连续，那么在点$M_{0}$满足
$\frac{\partial z}{\partial x\partial y}|_{(x_{0},y_{0})}=\frac{\partial z}{\partial y\partial x}|_{(x_{0},y_{0})}$
3、函数z=f(x,y)在点(x,y) 处的全增量$\Delta z=f(x+\Delta x,y+\Delta y)-f(x,y)$可以表示为$\Delta z=A\Delta x+B\Delta y+o(\rho )$。AB不依赖于$\Delta x,\Delta y$而仅仅与x,y有关，$\rho=\sqrt{\Delta x^{^{2}}+\Delta y^{^{2}}}$。
若z=f(x,y)在(x,y)可微，那么偏导数存在，且z=f(x,y)在点(x,y)可微。且$A=f^{{'}}_{x}(x,y),B=f^{{'}}_{y}(x,y)$。
证明：由于可微，所以$\Delta z=A\Delta x+B\Delta y+o(\rho )$。当$\Delta y=0$时，$\rho=|x_{0}|$，$\Delta z=A\Delta x+o(|x_{0}|)$，同时除以$\Delta x$，两端求极限
\[f_{x}^{^{'}}(x,y)=\lim_{\Delta x \rightarrow 0}\frac{\Delta z}{\Delta x}=\lim_{\Delta x\rightarrow 0}(A+\frac{o|\Delta x|}{\Delta x})=A\]
同理$B=f^{{'}}_{y}(x,y)$。
4、设函数z=f(x,y)在点$(x_{0},y_{0})$的某邻域内存在偏导数$f_{x}^{^{'}}(x,y),f_{y}^{^{'}}(x,y)$，并且$f_{x}^{^{'}}(x,y),f_{y}^{^{'}}(x,y)$都在点$(x_{0},y_{0})$连续，那么z=f(x,y)在点$(x_{0},y_{0})$可微。
证明：设
$\Delta z=f(x_{0}+\Delta x,y_{0}+\Delta y)-f(x_{0},y_{0})$
$=[f(x_{0}+\Delta x,y_{0}+\Delta y)-f(x_{0},y_{0}+\Delta y)]+[f(x_{0},y_{0}+\Delta y)-f(x_{0},y_{0}]$
将$f(x_{0}+\Delta x,y_{0}+\Delta y)-f(x_{0},y_{0}+\Delta y)$看做x的函数$f(x,y_{0}+\Delta y)$在$\Delta x$处的增量。由于$f_{x}^{^{'}}(x,y)$在$(x_{0},y_{0})$邻域内存在，所以$f(x,y_{0}+\Delta y)$在$\Delta X$某邻域内可导，根据微分中值定理，有
\[f(x_{0}+\Delta x,y_{0}+\Delta y)-f(x_{0},y_{0}+\Delta y)=f_{x}^{^{'}}(x_{0}+\theta_{1}\Delta x,y_{0}+\Delta y)\Delta x,(0<\theta_{1}<1)\]
同理
\[f(x_{0},y_{0}+\Delta y)-f(x_{0},y_{0})=f_{x}^{^{'}}(x_{0},y_{0}+\theta_{2} \Delta y)\Delta y,(0<\theta_{2}<1)\]
而$f_{x}^{^{'}}(x,y),f_{y}^{^{'}}(x,y)$都在点$(x_{0},y_{0})$连续所以
\[\lim_{\rho\rightarrow 0}f_{x}^{^{'}}(x_{0}+\theta_{1}\Delta x,y_{0}+\Delta y)=f_{x}^{^{'}}(x_{0},y_{0})\]
\[\lim_{\rho\rightarrow 0}f_{x}^{^{'}}(x_{0},y_{0}+\theta_{2}\Delta y)=f_{x}^{^{'}}(x_{0},y_{0})\]
所以存在无穷小$\alpha ,\beta $，当$\rho \rightarrow 0$时，有
\[f_{x}^{^{'}}(x_{0}+\theta_{1}\Delta x,y_{0}+\Delta y)=f_{x}^{^{'}}(x_{0},y_{0})+\alpha\]
\[f_{x}^{^{'}}(x_{0},y_{0}+\theta_{2}\Delta y)=f_{x}^{^{'}}(x_{0},y_{0})+\beta\]
所以
$\Delta z=f_{x}^{^{'}}(x_{0},y_{0})\Delta x+f_{y}^{^{'}}(x_{0},y_{0})\Delta y+\alpha\Delta x+\beta\Delta y$
由于
$|\frac{\alpha\Delta x+\beta\Delta y}{\rho}|$
$=|\frac{\alpha\Delta x+\beta\Delta y}{\sqrt{\Delta x^{^{2}}+\Delta y^{^{2}}}}|$
$\leq\frac{|\alpha\Delta x|}{\sqrt{\Delta x^{^{2}}+\Delta y^{^{2}}}}+\frac{|\beta\Delta y|}{\sqrt{\Delta x^{^{2}}+\Delta y^{^{2}}}}$
$\leq|\alpha|+|\beta|\rightarrow 0$
所以
$\lim_{\rho \rightarrow 0}\frac{\alpha\Delta x+\beta\Delta y}{\rho}=0$
即$\alpha\Delta x+\beta\Delta y=o(\rho)(\rho\rightarrow 0)$
所以$\Delta z=f_{x}^{^{'}}(x_{0},y_{0})\Delta x+f_{y}^{^{'}}(x_{0},y_{0})\Delta y+o(\rho)$。即z=f(x,y)在$(x_{0},y_{0})$可微。
5、设函数f(x,y)，$\varphi (x,y)$都具有连续偏导数，在$\varphi (x,y)=0$时求f(x,y)的极值。
(1)引入拉格朗日乘数$\lambda $,$F(x,y,\lambda)=f(x,y)+\lambda \varphi (x,y)$
(2)求三元函数$F(x,y,\lambda)$的驻点，即方程组
\[F_{x}^{^{'}}=f_{x}^{^{'}}(x,y)+\lambda \varphi _{x}^{^{'}}(x,y)=0\]
\[F_{y}^{^{'}}=f_{y}^{^{'}}(x,y)+\lambda \varphi _{y}^{^{'}}(x,y)=0\]
\[F_{\lambda}^{^{'}}=\varphi(x,y)=0\]
的所有解$(x_{0},y_{0},\lambda _{0})$
(3)判断$(x_{0},y_{0},\lambda _{0})$是否为$F(x,y,\lambda)$的极值点。