微积分学习笔记五：多元函数微积分

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 ightarrow 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$frac{partial z}{partial xpartial y}$和$frac{partial z}{partial ypartial x}$。如果$frac{partial z}{partial xpartial y}$和$frac{partial z}{partial ypartial x}$都在$M_{0}(x_{0},y_{0})$点连续，那么在点$M_{0}$满足
$frac{partial z}{partial xpartial y}|_{(x_{0},y_{0})}=frac{partial z}{partial ypartial x}|_{(x_{0},y_{0})}$

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$Delta z=ADelta x+BDelta y+o( ho )$。AB不依赖于$Delta x,Delta y$而仅仅与x,y有关，$ ho=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=ADelta x+BDelta y+o( ho )$。当$Delta y=0$时，$ ho=|x_{0}|$，$Delta z=ADelta x+o(|x_{0}|)$，同时除以$Delta x$，两端求极限
[f_{x}^{^{'}}(x,y)=lim_{Delta x ightarrow 0}frac{Delta z}{Delta x}=lim_{Delta x ightarrow 0}(A+frac{o|Delta x|}{Delta x})=A]
同理$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}+ heta_{1}Delta x,y_{0}+Delta y)Delta x,(0< heta_{1}<1)]
同理
[f(x_{0},y_{0}+Delta y)-f(x_{0},y_{0})=f_{x}^{^{'}}(x_{0},y_{0}+ heta_{2} Delta y)Delta y,(0< heta_{2}<1)]
而$f_{x}^{^{'}}(x,y),f_{y}^{^{'}}(x,y)$都在点$(x_{0},y_{0})$连续所以
[lim_{ ho ightarrow 0}f_{x}^{^{'}}(x_{0}+ heta_{1}Delta x,y_{0}+Delta y)=f_{x}^{^{'}}(x_{0},y_{0})]
[lim_{ ho ightarrow 0}f_{x}^{^{'}}(x_{0},y_{0}+ heta_{2}Delta y)=f_{x}^{^{'}}(x_{0},y_{0})]


所以存在无穷小$alpha ,eta $，当$ ho ightarrow 0$时，有
[f_{x}^{^{'}}(x_{0}+ heta_{1}Delta x,y_{0}+Delta y)=f_{x}^{^{'}}(x_{0},y_{0})+alpha]
[f_{x}^{^{'}}(x_{0},y_{0}+ heta_{2}Delta y)=f_{x}^{^{'}}(x_{0},y_{0})+eta]
所以
$Delta z=f_{x}^{^{'}}(x_{0},y_{0})Delta x+f_{y}^{^{'}}(x_{0},y_{0})Delta y+alphaDelta x+etaDelta y$
由于
$|frac{alphaDelta x+etaDelta y}{ ho}|$

$=|frac{alphaDelta x+etaDelta y}{sqrt{Delta x^{^{2}}+Delta y^{^{2}}}}|$
$leqfrac{|alphaDelta x|}{sqrt{Delta x^{^{2}}+Delta y^{^{2}}}}+frac{|etaDelta y|}{sqrt{Delta x^{^{2}}+Delta y^{^{2}}}}$

$leq|alpha|+|eta| ightarrow 0$
所以
$lim_{ ho ightarrow 0}frac{alphaDelta x+etaDelta y}{ ho}=0$
即$alphaDelta x+etaDelta y=o( ho)( ho ightarrow 0)$
所以$Delta z=f_{x}^{^{'}}(x_{0},y_{0})Delta x+f_{y}^{^{'}}(x_{0},y_{0})Delta y+o( ho)$。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。