【数学】多元函数微分学(宇哥笔记)

多元微分学

【数学】多元函数微分学(宇哥笔记)

概念

二重极限

$\begin{aligned} &若对\forall\epsilon>0,\exists\delta>0,当0<\sqrt{(x-x_0)^2+(y-y_0)^2}<\delta时，\\ &恒有|f(x,y)-A|<\epsilon,则称A是f(x,y)在(x_0,y_0)点的极限\\ [注](1)&\begin{cases}一元极限中x\to x_0有且仅有两种方式\\二元极限中有无穷任意多种方式\end{cases}\\ (2)&若有两条不同路径（如直线y=kx，抛物线x=y^2）使极限\lim_{{x\to0,y\to0}}f(x,y)值不相等\\ &或某一条路径使极限\lim_{{x\to0,y\to0}}f(x,y)值不存在，则说明二重极限\lim_{{x\to0,y\to0}}f(x,y)不存在\\ (3)&主要方法：化成一元极限、等价代换、无穷小乘有界、夹逼准则\\ (4)&二重极限保持了一元极限的各种性质，如唯一性、局部有界性、局部保号性及运算性质\\ (5)&所求极限得二元函数f(x,y)如果使齐次有理式函数，即分子、分母分别均是齐次有理函数，\\ &考察(x,y)\to(0,0)时得极限，可用下述命题：\\ &设f(x,y)=\frac{P(x,y)}{Q(x,y)}=\frac{m次}{n次},其中分子分母是互质多项式，则\\ &1.当m>n时，若方程Q(1,y)=0与Q(x,1)=0均无实根，则\lim_{{x\to0,y\to0}}f(x,y)=0;\\ &若方程Q(1,y)=0与Q(x,1)=0有实根，则\lim_{{x\to0,y\to0}}f(x,y)不存在\\ &2.当m\leq n时，\lim_{{x\to0,y\to0}}f(x,y)不存在\\ \end{aligned}$ [注](1)(2)(3)(4)(5)若对∀ϵ>0,∃δ>0,当0<(x−x0)2+(y−y0)2<δ时，恒有∣f(x,y)−A∣<ϵ,则称A是f(x,y)在(x0,y0)点的极限{一元极限中x→x0有且仅有两种方式二元极限中有无穷任意多种方式若有两条不同路径（如直线y=kx，抛物线x=y2）使极限x→0,y→0limf(x,y)值不相等或某一条路径使极限x→0,y→0limf(x,y)值不存在，则说明二重极限x→0,y→0limf(x,y)不存在主要方法：化成一元极限、等价代换、无穷小乘有界、夹逼准则二重极限保持了一元极限的各种性质，如唯一性、局部有界性、局部保号性及运算性质所求极限得二元函数f(x,y)如果使齐次有理式函数，即分子、分母分别均是齐次有理函数，考察(x,y)→(0,0)时得极限，可用下述命题：设f(x,y)=Q(x,y)P(x,y)=n次m次,其中分子分母是互质多项式，则1.当m>n时，若方程Q(1,y)=0与Q(x,1)=0均无实根，则x→0,y→0limf(x,y)=0;若方程Q(1,y)=0与Q(x,1)=0有实根，则x→0,y→0limf(x,y)不存在2.当m≤n时，x→0,y→0limf(x,y)不存在

$\begin{aligned} &\color{blue}[注3相关]\\ 1.&\color{maroon}\lim_{{x\to0,y\to0}}\frac{\sqrt{x^2+y^2}-\sin\sqrt{x^2+y^2}}{(\sqrt{x^2+y^2})^3}\\ &令\sqrt{x^2+y^2}=t,则I=\lim_{t\to0}\frac{t-\sin t}{t^3}=\frac16\\ 2.&\color{maroon}\lim_{{x\to0,y\to0}}\frac{\sin xy}{y}\\ &=\lim_{{x\to0,y\to0}}\frac{xy}{y}=\lim_{{x\to0,y\to0}}x=0\\ 3.&\color{maroon}\lim_{{x\to0,y\to0}}\frac{xy^2}{x^2+y^2}=\lim_{{x\to0,y\to0}}x\cdot\frac{y^2}{x^2+y^2}=0\\ &\color{blue}[注5相关]\\ 1.&\color{maroon}\lim_{{x\to0,y\to0}}\frac{x^3+y^3}{x^2+y^2}\\ & m=3>n=2,且Q(1,y)=1+y^2=0和Q(x,1)=x^2+1=0 无实根\implies I=0\\ 2.&\color{maroon}\lim_{{x\to0,y\to0}}\frac{xy}{x+y}\\ & m=2>n=1,且Q(1,y)=1+y=0有实根\implies 不\exists\\ 3.&\color{maroon}\lim_{{x\to0,y\to0}}\frac{x+y}{x-y}\\ & m=1=n=1\implies 不\exists\\ 4.&\color{maroon}\lim_{{x\to0,y\to0}}\frac{x^2+y^2}{x^3+y^3}\\ & m=2<n=3 \implies 不\exists\\ &\color{blue}[注2相关]\\ &\color{maroon}证明\lim_{{x\to0,y\to0}}\frac{x^2y}{x^4+y^2}不存在\\ &I\underrightarrow{y=kb}=\lim_{x\to0}\frac{kx^3}{x^4+k^2x^2}=\lim_{x\to0}\frac{kx}{x^2+k^2}=0\\ &I\underrightarrow{y=x^2}\lim_{x\to0}\frac{x^4}{x^4+x^4}=\frac12\\ &\therefore 二元极限不存在\\ \end{aligned}$ 1.2.3.1.2.3.4.[注3相关]x→0,y→0lim(x2+y2)3x2+y2−sinx2+y2令x2+y2=t,则I=t→0limt3t−sint=61x→0,y→0limysinxy=x→0,y→0limyxy=x→0,y→0limx=0x→0,y→0limx2+y2xy2=x→0,y→0limx⋅x2+y2y2=0[注5相关]x→0,y→0limx2+y2x3+y3m=3>n=2,且Q(1,y)=1+y2=0和Q(x,1)=x2+1=0无实根⟹I=0x→0,y→0limx+yxym=2>n=1,且Q(1,y)=1+y=0有实根⟹不∃x→0,y→0limx−yx+ym=1=n=1⟹不∃x→0,y→0limx3+y3x2+y2m=2<n=3⟹不∃[注2相关]证明x→0,y→0limx4+y2x2y不存在Iy=kb=x→0limx4+k2x2kx3=x→0limx2+k2kx=0Iy=x2x→0limx4+x4x4=21∴二元极限不存在

连续

$\begin{aligned} &若\lim_{{x\to0,y\to0}}f(x,y)=f(x_0,y_0),则称f(x,y)在点(x_0,y_0)处连续\\ &[注]不讨论间断点 \end{aligned}$

偏导数

$\begin{aligned} &z=f(x,y)在(x_0,y_0)处的偏导数\\ &f'_x(x_0,y_0)=\lim_{\Delta x\to0}\frac{f(x_0+\Delta x,y_0)-f(x_0,y_0)}{\Delta x}=\lim_{x\to x_0}\frac{f(x,y_0)-f(x_0,y_0)}{x-x_0}\\ &f'_y(x_0,y_0)=\lim_{\Delta y\to0}\frac{f(x_0,y_0+\Delta y)-f(x_0,y_0)}{\Delta y}=\lim_{y\to y_0}\frac{f(x_0,y)-f(x_0,y_0)}{y-y_0}\\ &1.偏导数实际上就是对应一元函数得导数，如:\\ &f_x'(x_0,y_0)=\varphi'(x)|_{x=x_0}=[f(x,y_0)]'|_{x\to x_0}\\ &f_y'(x_0,y_0)=\varphi'(y)|_{y=y_0}=[f(x_0,y)]'|_{y\to y_0}\\ &2.求f(x,y)的偏导数只需先把其中一个变量视为常数即可，\\ &如求f_x'(x,y)时，把f(x,y)中的y先视为常数，y偏导同理。 \end{aligned}$ z=f(x,y)在(x0,y0)处的偏导数fx′(x0,y0)=Δx→0limΔxf(x0+Δx,y0)−f(x0,y0)=x→x0limx−x0f(x,y0)−f(x0,y0)fy′(x0,y0)=Δy→0limΔyf(x0,y0+Δy)−f(x0,y0)=y→y0limy−y0f(x0,y)−f(x0,y0)1.偏导数实际上就是对应一元函数得导数，如:fx′(x0,y0)=φ′(x)∣x=x0=[f(x,y0)]′∣x→x0fy′(x0,y0)=φ′(y)∣y=y0=[f(x0,y)]′∣y→y02.求f(x,y)的偏导数只需先把其中一个变量视为常数即可，如求fx′(x,y)时，把f(x,y)中的y先视为常数，y偏导同理。

$\begin{aligned} 1.&\color{maroon}设f(x,y)=x^2+(y-1)\arcsin\sqrt{\frac yx},则\frac{\partial f}{\partial x}|_{(2,1)}=\\ &f(x,1)=x^2\implies \frac{\partial f}{\partial x}|_{(2,1)}=(x^2)'|_{x=2}=4\\ 2.&\color{maroon}设f(x,y)=\frac{e^x}{x-y},则\underline{\quad}\\ &f'_x=\frac{e^x\cdot(x-y)-e^x\cdot(1-0)}{(x-y)^2}\\ &f'_y=\frac{0-e^x\cdot(0-1)}{(x-y)^2}\\ &\implies f'_x+f'_y=\frac{e^x}{x-y}=f\\ 3.&\color{maroon}设f(x,y)=e^{x+y}[x^{\frac13}(y-1)^{\frac13}+y^{\frac13}(x-1)^{\frac23}],则在点(0,1)处的两个偏导数f_x'(0,1)=\underline{\quad},f_y'(0,1)=\underline{\quad}\\ &令y=1\implies f(x,1)=e^{x+1}(x-1)^{\frac23}\\ &\implies f_x'(x,1)=e^{x+1}(x-1)^\frac23+e^{x+1}\frac23(x-1)^{-\frac13}\\ &令x=0\implies f_x'(0,1)=e+e\cdot(-\frac23)=\frac e3\\ &f_y'(x_0,y_0)=\lim_{\Delta y\to0}\frac{f(x_0,y_0+\Delta y)-f(x_0,y_0)}{\Delta y}\\ &令x=0\implies f(0,y)=e^y\cdot y^{\frac13}\implies f_y'(0,y)=e^y\cdot y^{\frac13}+e^y\cdot\frac13y^{-\frac23}\\ &令y=1,f_y'(0,1)=e+e\cdot\frac13=\frac43e\\ \end{aligned}$ 1.2.3.设f(x,y)=x2+(y−1)arcsinxy,则∂x∂f∣(2,1)=f(x,1)=x2⟹∂x∂f∣(2,1)=(x2)′∣x=2=4设f(x,y)=x−yex,则fx′=(x−y)2ex⋅(x−y)−ex⋅(1−0)fy′=(x−y)20−ex⋅(0−1)⟹fx′+fy′=x−yex=f设f(x,y)=ex+y[x31(y−1)31+y31(x−1)32],则在点(0,1)处的两个偏导数fx′(0,1)=,fy′(0,1)=令y=1⟹f(x,1)=ex+1(x−1)32⟹fx′(x,1)=ex+1(x−1)32+ex+132(x−1)−31令x=0⟹fx′(0,1)=e+e⋅(−32)=3efy′(x0,y0)=Δy→0limΔyf(x0,y0+Δy)−f(x0,y0)令x=0⟹f(0,y)=ey⋅y31⟹fy′(0,y)=ey⋅y31+ey⋅31y−32令y=1,fy′(0,1)=e+e⋅31=34e

可微

$\begin{aligned} &判定二元函数f(x,y)在点(x_0,y_0)是否可微的方法\\ &先求f_x'(x_0,y_0)与f_y'(x_0,y_0),若有一个不存在，则直接不可微；\\ &若都存在，则检查\lim_{{x\to0,y\to0}}\frac{f(x,y)-f(x_0,y_0)-f_x'(x_0,y_0)(x-x_0)-f_y'(x_0,y_0)(y-y_0)}{\sqrt{(x-x_0)^2+（y-y_0)^2}}=0?\\ &若等于0，则可微，若不等于0或不存在，则不可微。\\ \end{aligned}$ 判定二元函数f(x,y)在点(x0,y0)是否可微的方法先求fx′(x0,y0)与fy′(x0,y0),若有一个不存在，则直接不可微；若都存在，则检查x→0,y→0lim(x−x0)2+（y−y0)2f(x,y)−f(x0,y0)−fx′(x0,y0)(x−x0)−fy′(x0,y0)(y−y0)=0?若等于0，则可微，若不等于0或不存在，则不可微。

$\begin{aligned} 1.&\color{maroon}讨论f(x,y)=\begin{cases}\frac{x^2y}{x^2+y^2},(x,y)\neq(0,0)\\0,(x,y)=(0,0)\end{cases}在(0,0)点可微性\\ &f_x'(0,0)=\lim_{x\to0}\frac{f(x,0)-f(0,0)}x=\lim_{x\to0}\frac{0-0}{x}=0(\exists)\\ &同理f_y'(0,0)=0(\exists)\\ &\lim_{{x\to0,y\to0}}\frac{f(x,y)-f(0,0)-0}{\sqrt{x^2+y^2}}=\lim_{{x\to0,y\to0}}\frac{x^2y}{(x^2+y^2)^{3/2}}\\ &(m=3=n=3)故f(x,y)在(0,0)处不可微\\ \end{aligned}$

概念之间的关系

两个偏导数连续→二元函数可微→二元函数连续→极限存在且二元函数可偏导

$\begin{aligned} 1.&\color{maroon}设f(x,y)具有一阶连续偏导数，且对任意的(x,y)都有\frac{\partial f(x.y)}{\partial x}>0,\frac{\partial f(x.y)}{\partial y}<0,则\\ &A.f(0,0)>f(1,1)\quad B.f(0,0)<f(1,1)\quad C.f(0,1)>f(1,0)\quad D.f(0,1)<f(1,0)\\ &由题意，得对x单调递增；对y单调递减\\ &画图可得D\\ 2.&\color{maroon}设f(x,y)=e^{\sqrt{x^2+y^4}},则\\ &f(x,0)=e^{|x|}在x=0处不可导\implies\frac{\partial f}{\partial x}|_{(0,0)}不\exists\\ &f(0,y)=e^{y^2}在y=0处可导\implies \frac{\partial f}{\partial y}|_{(0,0)}\exists\\ 3.&\color{maroon}设f(x,y)=\begin{cases}\frac{x^2y^2}{x^2+y^2},(x,y)\neq(0,0)\\0,(x,y)=(0,0)\end{cases},讨论f(x,y)在点(0,0)的连续性、可导性及可微性\\ &\lim_{{x\to0,y\to0}}f(x,y)=\lim_{{x\to0,y\to0}}\frac{x^2y^2}{x^2+y^2}=0=f(0,0),故连续\\ &f_x'(0,0)=\lim_{x\to0}\frac{f(x,0)-f(0,0)}x=\lim_{x\to0}\frac{0-0}{x}=0\quad\exists\\ &由对称性可知，f_y'(0,0)=0\quad\exists\\ &\lim_{{x\to0,y\to0}}\frac{f(x,y)-f(0,0)-0}{\sqrt{x^+y^2}}=\lim_{{x\to0,y\to0}}\frac{x^2y^2}{(x^2+y^2)^{3/2}}=0\\ &\implies f(x,y)在(0,0)处可微\\ 4.&\color{maroon}已知(axy^3-y^2\cos x)dx+(1+by\sin x+3x^2y^2)dy为某一函数u(x,y)的全微分，则\\ &\frac{\partial u}{\partial x}=axy^3-y^2\cos x\\ &\frac{\partial u}{\partial y}=1+by\sin x+3x^2y^2\\ &\implies \frac{\partial^2u}{\partial x\partial y}=3axy^2-2y\cos x\\ &\implies \frac{\partial^2u}{\partial y\partial x}=by\cos x+6xy^2\\ &\implies a=2,b=-2\\ \end{aligned}$ 1.2.3.4.设f(x,y)具有一阶连续偏导数，且对任意的(x,y)都有∂x∂f(x.y)>0,∂y∂f(x.y)<0,则A.f(0,0)>f(1,1)B.f(0,0)<f(1,1)C.f(0,1)>f(1,0)D.f(0,1)<f(1,0)由题意，得对x单调递增；对y单调递减画图可得D设f(x,y)=ex2+y4,则f(x,0)=e∣x∣在x=0处不可导⟹∂x∂f∣(0,0)不∃f(0,y)=ey2在y=0处可导⟹∂y∂f∣(0,0)∃设f(x,y)={x2+y2x2y2,(x,y)̸=(0,0)0,(x,y)=(0,0),讨论f(x,y)在点(0,0)的连续性、可导性及可微性x→0,y→0limf(x,y)=x→0,y→0limx2+y2x2y2=0=f(0,0),故连续fx′(0,0)=x→0limxf(x,0)−f(0,0)=x→0limx0−0=0∃由对称性可知，fy′(0,0)=0∃x→0,y→0limx+y2f(x,y)−f(0,0)−0=x→0,y→0lim(x2+y2)3/2x2y2=0⟹f(x,y)在(0,0)处可微已知(axy3−y2cosx)dx+(1+bysinx+3x2y2)dy为某一函数u(x,y)的全微分，则∂x∂u=axy3−y2cosx∂y∂u=1+bysinx+3x2y2⟹∂x∂y∂2u=3axy2−2ycosx⟹∂y∂x∂2u=bycosx+6xy2⟹a=2,b=−2

计算

链式求导规则+高阶偏导复合结构不变

$\begin{aligned} 1.&\color{maroon}设z=f(u,v,x),u=u(x,y),v=v(y)都是可微函数，求\frac{\partial z}{\partial x}和\frac{\partial z}{\partial y}\\ &\frac{\partial z}{\partial x}=f_1'\cdot\frac{\partial u}{\partial x}+f_3'\cdot1\\ &\frac{\partial z}{\partial y}=f_1'\frac{\partial u}{\partial y}+f_2'\cdot\frac{dv}{dy}\\ &\color{grey}[注]u是二元\to \partial,v是一元\to d\\ 2.&\color{maroon}设z=f(e^x\sin y,x^2+y^2),其中f具有二阶连续偏导数，求\frac{\partial^2 z}{\partial x\partial y}\\ &\frac{\partial z}{\partial x}=f_1'\cdot e^x\sin y+f_2'\cdot2x\\ &\frac{\partial^2z}{\partial x \partial y}=(f_{11}''\cdot e^x\cos y+f_{12}''\cdot2y)\cdot e^x\sin y+f_1'\cdot e^x\cos y+2x(f_{21}''\cdot e^x\cos y+f_{22}''\cdot2y)\\ &=e^{2x}\sin y\cos y\cdot f_{11}''+(2ye^x\sin y+2xe^x\cos y)\cdot f_{12}''+e^x\cos y f_1'+4xyf_{22}''\\ 3.&\color{maroon}设z=f(u,v,x),u=xe^y,其中f具有二阶连续偏导数，求\frac{\partial^2z}{\partial x\partial y}\\ &\frac{\partial z}{\partial x}=f_1'\cdot e^y+f_2'\cdot1\\ &\frac{\partial^2z}{\partial x \partial y}=(f_{11}''\cdot xe^y+f_{13}''\cdot1)\cdot e^y+f_1'\cdot e^y+(f_{21}''\cdot xe^y+f_{23}''\cdot1)\\ 4.&\color{maroon}设z=f(2x-y)+g(x,xy),其中f二阶可导，g具有二阶连续偏导数，求\frac{\partial^2z}{\partial x \partial y}\\ &\frac{\partial z}{\partial x}=f'\cdot2+g_1'\cdot1+g_2'\cdot y\\ &\frac{\partial^2z}{\partial x\partial y}=2f''\cdot(-1)+g_{12}''\cdot x+g_{22}''\cdot x\cdot y+g_2'\cdot1\\ 5.&\color{maroon}设F(u,v)二阶偏导连续，并设z=F(\frac yx,x^2+y^2),求\frac{\partial^2z}{\partial x\partial y}\\ &\frac{\partial z}{\partial x}=F_1'(-\frac{y}{x^2})+F_2'\cdot 2x\\ &\frac{\partial^2z}{\partial x\partial y}=\frac{\partial(\frac{\partial z}{\partial x})}{\partial y}=\frac{\partial(F_1'\cdot(-\frac y{x^2}))}{\partial y}+\frac{\partial(F_2'\cdot2x)}{\partial y}\\ &=\frac{\partial F_1'}{\partial y}(-\frac y{x^2})+F_1'(-\frac1{x^2})+2x\frac{\partial F_2'}{\partial y}\\ &=(F_{11}''(\frac1x)+F_{12}''2y)(-\frac y{x^2})+F_1'(-\frac1{x^2})+2x(F_{21}''(\frac1x)+F_{22}''2y)\\ \end{aligned}$ 1.2.3.4.5.设z=f(u,v,x),u=u(x,y),v=v(y)都是可微函数，求∂x∂z和∂y∂z∂x∂z=f1′⋅∂x∂u+f3′⋅1∂y∂z=f1′∂y∂u+f2′⋅dydv[注]u是二元→∂,v是一元→d设z=f(exsiny,x2+y2),其中f具有二阶连续偏导数，求∂x∂y∂2z∂x∂z=f1′⋅exsiny+f2′⋅2x∂x∂y∂2z=(f11′′⋅excosy+f12′′⋅2y)⋅exsiny+f1′⋅excosy+2x(f21′′⋅excosy+f22′′⋅2y)=e2xsinycosy⋅f11′′+(2yexsiny+2xexcosy)⋅f12′′+excosyf1′+4xyf22′′设z=f(u,v,x),u=xey,其中f具有二阶连续偏导数，求∂x∂y∂2z∂x∂z=f1′⋅ey+f2′⋅1∂x∂y∂2z=(f11′′⋅xey+f13′′⋅1)⋅ey+f1′⋅ey+(f21′′⋅xey+f23′′⋅1)设z=f(2x−y)+g(x,xy),其中f二阶可导，g具有二阶连续偏导数，求∂x∂y∂2z∂x∂z=f′⋅2+g1′⋅1+g2′⋅y∂x∂y∂2z=2f′′⋅(−1)+g12′′⋅x+g22′′⋅x⋅y+g2′⋅1设F(u,v)二阶偏导连续，并设z=F(xy,x2+y2),求∂x∂y∂2z∂x∂z=F1′(−x2y)+F2′⋅2x∂x∂y∂2z=∂y∂(∂x∂z)=∂y∂(F1′⋅(−x2y))+∂y∂(F2′⋅2x)=∂y∂F1′(−x2y)+F1′(−x21)+2x∂y∂F2′=(F11′′(x1)+F12′′2y)(−x2y)+F1′(−x21)+2x(F21′′(x1)+F22′′2y)

隐函数求导

$\begin{aligned} &F(x,y,z)=0\implies z=z(x,y)\implies F(x,y,z(x,y))=0 \end{aligned}$

$\begin{aligned} &\frac{\partial F}{\partial x}=F_x'=F_1'+F_z'\frac{\partial z}{\partial x}=0\implies\begin{cases}\frac{\partial z}{\partial x}=-\frac{F_x'}{F_z'}\\\frac{\partial z}{\partial y}=-\frac{F_y'}{F_z'}\end{cases} \end{aligned}$

$\begin{aligned} 1.&\color{maroon}设z=z(x,y)由方程\ln z+e^{z-1}=xy确定，则\frac{\partial z}{\partial x}|_{(2,\frac12)}\\ 方法一.&由\ln z+e^{z-1}=xy\implies \frac1z\cdot z_x'+e^{z-1}\cdot z_x'=y\\ &由x=2,y=\frac12,代入前者，则z=1;代入后者，则z_x'=\frac14\\ 方法二.&令F(x,y,z)=\ln z+e^{z-1}-xy=o\\ &\implies\frac{\partial z}{\partial x}=-\frac{F_x'}{F_z'}=-\frac{-y}{\frac1z+e^{z-1}}|_{(2,\frac1x,1)}=\frac14\\ 2.&\color{maroon}设z=z(x,y)由方程F(x+\frac zy,y+\frac zx)=0确定，其中F由连续偏导数，求x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}\\ 方法一.&F_1'\cdot(1+\frac1y\frac{\partial z}{\partial x})+F_2'\cdot\frac{\frac{\partial z}{\partial x}\cdot x-z}{x^2}=0(对x)\\ &F_1'\cdot\frac{\frac{\partial z}{\partial y}\cdot y-z}{y^2}+F_2'\cdot(1+\frac1x\cdot\frac{\partial z}{\partial y})=0(对y)\\ &\implies x\cdot\frac{\partial z}{\partial x}+y\cdot\frac{\partial z}{\partial y}=z-xy\\ 方法二.&\frac{\partial z}{\partial x}=-\frac{F_x'}{F_z'}=-\frac{F_1'+F_2'(-\frac{z}{x^2})}{F_1'(\frac1y)+F_2'(\frac1x)}\\ &\frac{\partial z}{\partial y}=-\frac{F_y'}{F_z'}=-\frac{F_1'(-\frac{z}{y^2})+F_2'}{F_1'(\frac1y)+F_2'(\frac1x)}\\ &I=z-xy\\ 3.&\color{maroon}设\begin{cases}u=f(x-ut,y-ut,z-ut)\\g(x,y,z)\end{cases},求\frac{\partial u}{\partial x},\frac{\partial u}{\partial y}\\ &[分析]一般有几个方程，就有几个因变量，其余的字母都是自变量\\ &\frac{\partial u}{\partial x}=f_1'\cdot(1-\frac{\partial u}{\partial x}t)+f_2'\cdot(-\frac{\partial u}{\partial x}t)+f_3'\cdot(\frac{\partial z}{\partial x}-\frac{\partial u}{\partial x}t)\\ &g_1'\cdot1+g_2'\cdot0+g_3'\cdot\frac{\partial z}{\partial x}=0\\ &解得\frac{\partial u}{\partial x}=\frac{f_1'g_3'-f_3'g_1'}{g_3'[1+t(f_1'+f_2'+f_3')]}\\ &对y求偏导数同样可得\frac{\partial u}{\partial y}=\frac{f_2'g_3'-f_3'g_2'}{g_3'[1+t(f_1'+f_2'+f_3')]}\\ \end{aligned}$ 1.方法一.方法二.2.方法一.方法二.3.设z=z(x,y)由方程lnz+ez−1=xy确定，则∂x∂z∣(2,21)由lnz+ez−1=xy⟹z1⋅zx′+ez−1⋅zx′=y由x=2,y=21,代入前者，则z=1;代入后者，则zx′=41令F(x,y,z)=lnz+ez−1−xy=o⟹∂x∂z=−Fz′Fx′=−z1+ez−1−y∣(2,x1,1)=41设z=z(x,y)由方程F(x+yz,y+xz)=0确定，其中F由连续偏导数，求x∂x∂z+y∂y∂zF1′⋅(1+y1∂x∂z)+F2′⋅x2∂x∂z⋅x−z=0(对x)F1′⋅y2∂y∂z⋅y−z+F2′⋅(1+x1⋅∂y∂z)=0(对y)⟹x⋅∂x∂z+y⋅∂y∂z=z−xy∂x∂z=−Fz′Fx′=−F1′(y1)+F2′(x1)F1′+F2′(−x2z)∂y∂z=−Fz′Fy′=−F1′(y1)+F2′(x1)F1′(−y2z)+F2′I=z−xy设{u=f(x−ut,y−ut,z−ut)g(x,y,z),求∂x∂u,∂y∂u[分析]一般有几个方程，就有几个因变量，其余的字母都是自变量∂x∂u=f1′⋅(1−∂x∂ut)+f2′⋅(−∂x∂ut)+f3′⋅(∂x∂z−∂x∂ut)g1′⋅1+g2′⋅0+g3′⋅∂x∂z=0解得∂x∂u=g3′[1+t(f1′+f2′+f3′)]f1′g3′−f3′g1′对y求偏导数同样可得∂y∂u=g3′[1+t(f1′+f2′+f3′)]f2′g3′−f3′g2′

应用

无条件极值

$\begin{aligned} 1.&必要条件\begin{cases}f_x'(x_0,y_0)=0\\f_y'(x_0,y_0)=0\end{cases}\\ 2.&充分条件\begin{cases}A>0且AC-B^2>0\implies极小\\A<0且AC-B^2>0\implies极大\\AC-B^2<0\implies非极值点\end{cases}\\ [注]&记A=f_{xx}''(x_0,y_0),B=f_{xy}''(x_0,y_0),C=f_{yy}''(x_0,y_0)\\ \end{aligned}$

条件极值

$\begin{aligned} 1.&构造拉格朗日函数F(x,y,\lambda)=f(x,y)+\lambda\varphi(x,y),\\ 2.&令\begin{cases}F_x'=f_x'(x,y)+\lambda\varphi_x'(x,y)=0\\F_y'=f_y'(x,y)+\lambda\varphi_y'(x,y)=0\\F_\lambda'=\varphi(x,y)=0\end{cases}\\ 3.&比较上述各函数值的大小，最大的为最大值，最小的为最小值\\ \end{aligned}$

$\begin{aligned} 1.&\color{maroon}设z=z(x,y)由方程(x^2+y^2)z+\ln z+2(x+y+1)=0确定，求z=z(x,y)的极值\\ &由方程得2xz+(x^2+y^2)z_x'+\frac1zz_x'+2=0\\ &\implies2yz+(x^2+y^2)z_y'+\frac1zz_y'+2=0\\ &令z_x'=0,z_y'=0,则x=y=-\frac1z代入第一个式子\\ &得：\frac2z+\ln z-\frac4z+2=0\implies z=1\\ &\therefore x=y=-1,z=1\\ &\implies 2z+2xz_x'+2xz_x'+(x^2+y^2)z_{xx}''+\frac{z_{xx}''\cdot z-z_x'^2}{z^2}=0\\ &\implies 2xz_y'+2yz_x'+(x^2+y^2)z_{xy}''+\frac{z_{xy}''\cdot z-z_x'\cdot z_y'}{z^2}=0\\ &A=z_{xx}''(-1,-1)=-\frac23=c(对称性)\\ &B=z_{xy}''(-1,-1)=0,由B^2-AC<0且A<0\\ &故z(-1,-1)=1极大值\\ 2.&\color{maroon}求函数u=x^2+y^2+z^2在条件z=x^2+y^2及x+y+z=4下的最大值与最小值\\ &令F(x,y,z,\lambda,\mu)=x^2+y^2+z^2+\lambda(x^2+y^2-z)+\mu(x+y+z-4)\\ 得&\begin{cases}F_x'=2x+2\lambda x+\mu=0\\ F_y'=2y+2xy+\mu=0\\ F_z'=2z-\lambda+\mu=0\\ F_\lambda'=x^2+y^2-z=0\\ F_\mu'=x+y+z-4=0\end{cases}解得：P_1(1,1,2),P_2(-2,-2,8)\\ &由u(P_1)=6为最小值,且u(P_2)=72为最大值\\ 3.&\color{maroon}求u=\sqrt{x^2+y^2+z^2}在(x-y)^2-z^2=1条件下的最小值\\ &令F(x,y,z,\lambda)=x^2+y^2+z^2+\lambda((x-y)^2-z^2-1)\\ &\begin{cases}F_x'=2x+2\lambda(x-y)=0\\ F_y'=2y-2\lambda(x-y)=0\\ F_z'=2z-2\lambda z=0\\ F_\lambda'=(x-y)^2-z^2-1=0\end{cases}解得：P_1(-\frac12,\frac12,0),P_2(\frac12,-\frac12,0)\\ &由u(P_1)=u(P_2)=\frac{\sqrt{2}}2\\ 4.&\color{maroon}求f(x,y)=x^2-y^2+2在椭圆域D:x^2+\frac{y^2}4\leq1上的最大值与最小值\\ &内部\to f(x,y),由f_x'=2x=0,f_y'=-2y=0\\ &边界\to F(x,y,\lambda)\\ &F(x,y,\lambda)=x^2-y^2+\lambda(x^2+\frac{y^2}4-1)\\ &由F_x'=2x+2\lambda x=0,F_y'=-2y+\frac{\lambda}2y=0\\ &F_x'=x^2+\frac{y^2}4-1=0\\ &f(0,0)=2,f(+-1,0=3)最大,f(0,+-2)=-2最小\\ 5.&\color{maroon}求f(x,y)=x+xy-x^2-y^2在闭区域D:0\leq x\leq1,0\leq y\leq2上得最大值与最小值\\ &内部\to f(x,y),由f_x'=1+y-2x=0,f_y'=x-2y=0\\ &边界(代入法)L_1:y=0(0\leq x\leq1)\implies f(x,0)=x-x^2=\varphi(x)\implies (\frac12,0),(0,0),(1,0)\\ &L_2:x=1(0<y<2) \implies f(1,y)=y-y^2\implies (1,\frac12)\\ &L_3:y=2(0\leq x\leq1)\implies f(x,2)=3x-x^2-4\implies (0,2),(1,2)\\ &L_4:x=0(0<y<2) \implies f(0,y)=-y^2\\ &比较得最大值为\frac13，最小值为-4 \end{aligned}$ 1.2.得3.4.5.设z=z(x,y)由方程(x2+y2)z+lnz+2(x+y+1)=0确定，求z=z(x,y)的极值由方程得2xz+(x2+y2)zx′+z1zx′+2=0⟹2yz+(x2+y2)zy′+z1zy′+2=0令zx′=0,zy′=0,则x=y=−z1代入第一个式子得：z2+lnz−z4+2=0⟹z=1∴x=y=−1,z=1⟹2z+2xzx′+2xzx′+(x2+y2)zxx′′+z2zxx′′⋅z−zx′2=0⟹2xzy′+2yzx′+(x2+y2)zxy′′+z2zxy′′⋅z−zx′⋅zy′=0A=zxx′′(−1,−1)=−32=c(对称性)B=zxy′′(−1,−1)=0,由B2−AC<0且A<0故z(−1,−1)=1极大值求函数u=x2+y2+z2在条件z=x2+y2及x+y+z=4下的最大值与最小值令F(x,y,z,λ,μ)=x2+y2+z2+λ(x2+y2−z)+μ(x+y+z−4)⎩⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎧Fx′=2x+2λx+μ=0Fy′=2y+2xy+μ=0Fz′=2z−λ+μ=0Fλ′=x2+y2−z=0Fμ′=x+y+z−4=0解得：P1(1,1,2),P2(−2,−2,8)由u(P1)=6为最小值,且u(P2)=72为最大值求u=x2+y2+z2在(x−y)2−z2=1条件下的最小值令F(x,y,z,λ)=x2+y2+z2+λ((x−y)2−z2−1)⎩⎪⎪⎪⎨⎪⎪⎪⎧Fx′=2x+2λ(x−y)=0Fy′=2y−2λ(x−y)=0Fz′=2z−2λz=0Fλ′=(x−y)2−z2−1=0解得：P1(−21,21,0),P2(21,−21,0)由u(P1)=u(P2)=22求f(x,y)=x2−y2+2在椭圆域D:x2+4y2≤1上的最大值与最小值内部→f(x,y),由fx′=2x=0,fy′=−2y=0边界→F(x,y,λ)F(x,y,λ)=x2−y2+λ(x2+4y2−1)由Fx′=2x+2λx=0,Fy′=−2y+2λy=0Fx′=x2+4y2−1=0f(0,0)=2,f(+−1,0=3)最大,f(0,+−2)=−2最小求f(x,y)=x+xy−x2−y2在闭区域D:0≤x≤1,0≤y≤2上得最大值与最小值内部→f(x,y),由fx′=1+y−2x=0,fy′=x−2y=0边界(代入法)L1:y=0(0≤x≤1)⟹f(x,0)=x−x2=φ(x)⟹(21,0),(0,0),(1,0)L2:x=1(0<y<2)⟹f(1,y)=y−y2⟹(1,21)L3:y=2(0≤x≤1)⟹f(x,2)=3x−x2−4⟹(0,2),(1,2)L4:x=0(0<y<2)⟹f(0,y)=−y2比较得最大值为31，最小值为−4