Section02_基本理论

有界性和介值

  • 若 f(x,y)∈C(D)f(x,y)\in C(\mathbb{D})，D\mathbb{D} 为有界闭区间，则
    1. Th1 ∃ m,M\exists\ m,M
    2. Th2 ∃ k>0\exists\ k>0，使得 ∣f(x,y)∣≤k\vert f(x,y)\vert\le k
    3. Th3 ∀ δ∈[m,M], ∃ (ξ,η)∈D, f(ξ,η)=δ\forall\ \delta\in[m,M],\ \exists\ (\xi,\eta)\in\mathbb{D},\ f(\xi,\eta) = \delta

连续性，可偏导性，可微性的关系

graph TD id1(连续可偏导) id2(可微) id3.1(连续性) id3.2(可偏导性) id4[强] id5[弱] id1 --> id2 --> id3.1 & id3.2 id4 ---> id5

Proof

可微 ==> 连续 ∵Δz=f(x,y)−f(x0,y0)=A(x−x0)+B(y−y0)+o(ρ)∴lim⁡x→x0y→y0Δz=lim⁡x→x0y→y0f(x,y)−f(x0,y0)=0∴lim⁡x→x0y→y0f(x,y)=f(x0,y0)\begin{array}{ll} \because & \Delta z = f(x, y)- f(x_{0}, y_{0}) = A(x-x_{0}) + B(y-y_{0}) + o(\rho) \\ \therefore &\displaystyle \lim_{\substack{x\to x_{0}\\ y \to y_{0}}} \Delta z = \lim_{\substack{x\to x_{0}\\y\to y_{0}}}f(x,y) - f(x_{0},y_{0}) = 0 \\ \therefore & \displaystyle \lim_{\substack{x\to x_{0}\\ y\to y_{0}}}f(x,y) = f(x_{0},y_{0}) \\\end{array} 可微 ==> 可偏导 ∵Δz=AΔx+BΔy+o(ρ)取Δy=0∴lim⁡Δx→0ΔzxΔx=A,存在对x的偏导数同理可得: lim⁡Δy→0ΔzyΔy=B,存在对于y的偏导数\begin{array}{ll} \because & \Delta z = A \Delta x + B\Delta y + o(\rho)\\ &\text{取}\Delta y = 0 \\ \therefore & \displaystyle \lim_{\Delta x \to 0}\frac{\Delta z_{x}}{\Delta x} = A, \text{存在对}x \text{的偏导数} \\ & \displaystyle\text{同理可得: } \lim_{\Delta y\to 0} \frac{\Delta z_{y}}{\Delta y} =B, \text{存在对于}y \text{的偏导数}\end{array}

反例

连续不一定可偏导 f(x,y)=x2+y2f(x,y) = \sqrt{x^{2}+y^{2}} f(x,y)=x2+y2在(x,y)=(0,0)处连续而∂z∂x=lim⁡x→0x2x∵lim⁡x→0−x2x=−1≠lim⁡x→0+x2x=1∴f(x,y)在(0,0)对x不可偏导\begin{array}{ll} & f(x,y) = \sqrt{x^{2} + y^{2}} \text{在} (x,y) = (0,0) \text{处连续}\\ & \displaystyle\text{而} \frac{\partial z}{\partial x} = \lim_{x\to 0} \frac{\sqrt{x^{2}}}{x} \\ \because & \displaystyle \lim_{x\to 0^{-}} \frac{\sqrt{x^{2}}}{x} = -1 \ne \lim_{x\to 0^{+}} \frac{\sqrt{x^{2}}}{x} =1\\ \therefore & f(x,y)\text{在}(0,0)\text{对}x \text{不可偏导} \\\end{array} 可偏导不一定连续 f(x,y)={xyx2+y2(x,y)≠(0,0)0(x,y)=(0,0)f(x,y) = \begin{cases}\frac{xy}{x^{2}+y^{2}}&(x,y)\ne(0,0)\\0&(x,y)=(0,0)\end{cases} ∵lim⁡x→0y=xxyx2+y2=12≠lim⁡x→0y=−x=−12∴f(x,y)在(0,0)不连续∵lim⁡x→0f(x)−f(0)x−0=lim⁡x→00x3=0同理可得 fy(0,0)=0∴f(x,y)可偏导\begin{array}{ll} \because & \displaystyle \lim_{\substack{x\to 0\\y=x}} \frac{xy}{x^{2}+y^{2}} = \frac{1}{2} \ne \lim_{\substack{x\to 0\\y=-x}} = -\frac{1}{2} \\ \therefore & f(x,y) \text{在}(0,0)\text{不连续} \\ \because & \displaystyle \lim_{x\to 0} \frac{f(x)-f(0)}{x-0} = \lim_{x\to 0} \frac{0}{x^{3}} = 0\\ & \text{同理可得 } \displaystyle f_{y}(0,0) = 0 \\ \therefore & f(x,y)\text{可偏导} \\\end{array}

二阶连续可偏导

  • 若 z=f(x,y)z=f(x,y) 二阶连续可偏导，则 ∂2z∂x∂y=∂2z∂y∂x \frac{\partial^{2} z}{\partial x\partial y} = \frac{\partial^{2} z}{\partial y\partial x}

求偏导类型

显函数求偏导 z=f(x,y)z=f(x,y)

  1. z=ln⁡2arctan⁡x+yx−y\displaystyle z = \ln^{2}\arctan \frac{x+y}{x-y} ∂z∂x=2ln⁡arctan⁡x+yx−y⋅1arctan⁡x+yx−y⋅11+(x+yx−y)2⋅(x−y)−(x+y)(x−y)2 \begin{array}{ll} & \displaystyle \frac{\partial z}{\partial x} = 2 \ln \arctan \frac{x+y}{x-y} \cdot \frac{1}{\arctan \frac{x+y}{x-y}} \cdot \frac{1}{1+\big(\frac{x+y}{x-y}\big)^{2}}\cdot \frac{(x-y)- (x+y)}{(x-y)^{2}} \end{array}
  2. z=(1+x2+y2)xy\displaystyle z= (1+x^{2}+y^{2})^{xy} ∵z=(1+x2+y2)xy=exyln⁡(1+x2+y2)∴∂z∂x=exyln⁡(1+x2+y2)⋅[yln⁡(1+x2+y2)+xy2x1+x2+y2]=(1+x2+y2)xy⋅[yln⁡(1+x2+y2)+xy2x1+x2+y2]\begin{array}{ll} \because & \displaystyle z= (1+x^{2}+y^{2})^{xy} = e^{xy\ln(1+x^{2}+y^{2})}\\ \therefore & \displaystyle \frac{\partial z}{\partial x} = e^{xy\ln(1+x^{2}+y^{2})} \cdot [y\ln(1+x^{2}+y^{2})+xy \frac{2x}{1+x^{2}+y^{2}}] \\ & = (1+x^{2}+y^{2})^{xy}\cdot [y\ln(1+x^{2}+y^{2})+xy \frac{2x}{1+x^{2}+y^{2}}]\end{array}

复合函数求导

分析实例

  1. z=f(x2+y2)⇒z=f(u),u=x2+y2z=f(x^{2}+y^{2})\Rightarrow z=f(u), u = x^{2}+y^{2}
    1. zz 为关于 x,yx,y 的二元函数
    2. f()f() 为一元函数
  2. z=f(t2,et)⇒z=f(u,v),{u=t2v=etz=f(t^{2},e^{t})\Rightarrow z = f(u,v),\begin{cases}u = t^{2}\\v=e^t\end{cases}
    1. zz 为关于 tt 的一元函数
    2. f(u,v)f(u,v) 为二元函数
  3. z=f(x2+y2,xy,x)⇒z=f(u,v,w),{u=x2+y2v=xyw=yz=f(x^{2}+y^{2},xy,x)\Rightarrow z=f(u,v,w),\begin{cases}u=x^{2}+y^{2}\\v=xy\\w=y\end{cases}
    1. zz 为关于 x,yx,y 的二元函数
    2. f(u,v,w)f(u,v,w) 为三元函数

表示方式 (对于z=f(u,v)z=f(u,v)

∂z∂u=fu(u,v)=f1(u,v)=f1\displaystyle \frac{\partial z}{\partial u} = f_{u}(u,v) = f_{1}(u,v)=f_{1} ∂z∂v=fv(u,v)=f2(u,v)=f2\displaystyle \frac{\partial z}{\partial v} = f_{v}(u,v) = f_{2}(u,v)=f_{2} ∂2z∂u2=fuu(u,v)=f11(u,v)=f11\displaystyle \frac{\partial^{2} z}{\partial u^{2}} = f_{uu}(u,v) = f_{11}(u,v)=f_{11} ∂2z∂u∂v=fuv(u,v)=f12(u,v)=f12\displaystyle \frac{\partial^{2} z}{\partial u\partial v} = f_{uv}(u,v) = f_{12}(u,v)=f_{12}

  1. z=eu2+sin⁡v,{u=xyv=yx\displaystyle z=e^{u^{2}+\sin v},\begin{cases}u=xy\\v=\frac{y}{x}\end{cases} 求 ∂z∂x,∂z∂y\displaystyle \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} Method 1z=ex2y2+sin⁡yx∴∂z∂x=ex2y2+sin⁡yx⋅(2y2x−yx2cos⁡yx)∂z∂y=ex2y2+sin⁡yx⋅(2x2y+1xcos⁡yx)Method 2∂z∂x=∂z∂u∂u∂x+∂z∂v∂v∂x=2uyeu2+sin⁡v−yx2eu2+sin⁡vcos⁡v=eu2+sin⁡v(2uy−yx2cos⁡v)=ex2y2+sin⁡yx(2y2x−yx2cos⁡yx) \begin{array}{ll} & \text{Method 1} \\ & \displaystyle z= e^{x^{2}y^{2}+\sin \frac{y}{x}}\\ \therefore & \displaystyle \frac{\partial{z}}{\partial{x}} = e^{x^{2}y^{2}+\sin \frac{y}{x}}\cdot \bigg(2y^{2}x - \frac{y}{x^{2}}\cos \frac{y}{x} \bigg) \\ & \displaystyle \frac{\partial{z}}{\partial{y}} = e^{x^{2}y^{2}+\sin \frac{y}{x}}\cdot \bigg(2x^{2}y + \frac{1}{x}\cos \frac{y}{x} \bigg) \\\\ & \color{#D0104C}\text{Method 2}\\ & \displaystyle \frac{\partial{z}}{\partial{x}} = \frac{\partial{z}}{\partial{u}}\frac{\partial{u}}{\partial{x}} + \frac{\partial{z}}{\partial{v}}\frac{\partial{v}}{\partial{x}} = 2uye^{u^{2}+\sin v} - \frac{y}{x^{2}}e^{u^{2}+\sin v}\cos v \\ &\displaystyle = e^{u^{2}+\sin v}\bigg(2uy - \frac{y}{x^{2}}\cos v\bigg) = e^{x^{2}y^{2}+\sin \frac{y}{x}}\bigg(2y^{2}x - \frac{y}{x^{2}}\cos \frac{y}{x}\bigg) \\ \end{array}

  2. z=f(t2,et),f(u,v)\displaystyle z=f(t^{2},e^{t}), f(u,v) 二阶连续可偏导，求d2zdt2\displaystyle \frac{d^{2}{z}}{d{t^{2}}} dzdt=∂z∂u⋅∂u∂t+∂z∂v⋅∂v∂t=2tf1+etf2d2zdt2=2t(2tf11+etf12)+2f1+et(2tf21+etf22)+etf2=2f1+etf2+4t2f11+4tetf12+e2tf22 \begin{array}{ll} & \displaystyle \frac{d{z}}{d{t}} = \frac{\partial{z}}{\partial{u}}\cdot \frac{\partial{u}}{\partial{t}} + \frac{\partial{z}}{\partial{v}}\cdot \frac{\partial{v}}{\partial{t}} = 2tf_{1} + e^{t}f_{2} \\ & \displaystyle \frac{d^{2}z}{dt^{2}} = 2t(2tf_{11}+e^{t}f_{12})+2f_{1} + e^{t}(2tf_{21}+e^{t}f_{22}) + e^{t}f_{2}\\ & \displaystyle = 2f_{1} + e^{t}f_{2} + 4t^{2}f_{11}+4te^{t}f_{12}+e^{2t}f_{22} \end{array}

  3. z=f(x2+y2,exsin⁡y),f(u,v)\displaystyle z= f(x^{2} + y^{2}, e^{x}\sin y), f(u,v) 二阶连续可偏导，求 ∂2z∂x∂y\displaystyle \frac{\partial^{2} z}{\partial x\partial y} ∂z∂x=2xf1+exf2sin⁡y∂2z∂x∂y=2x(2yf11+exf12cos⁡y)+exf2cos⁡y+ex(2yf21+exf22cos⁡y)sin⁡y=4xyf11+2ex(xcos⁡y+ysin⁡y)f12+excos⁡yf2+e2xcos⁡ysin⁡yf22 \begin{array}{ll} & \displaystyle \frac{\partial{z}}{\partial{x}} = 2x f_{1} + e^{x} f_{2} \sin y \\ & \displaystyle \frac{\partial^{2}{z}}{\partial{x}\partial{y}} = 2x(2yf_{11} + e^{x}f_{12}\cos y) + e^{x} f_{2} \cos y + e^{x}(2yf_{21} + e^{x}f_{22}\cos y)\sin y \\ & = 4xy f_{11} + 2e^{x}(x\cos y+y\sin y) f_{12} + e^{x}\cos y f_{2} + e^{2x}\cos y \sin y f_{22} \end{array}

  4. z=f(x+y,xy,x2),f(u,v,w)\displaystyle z= f(x+y,xy,x^{2}),f(u,v,w) 二阶连续可偏导，求 ∂2z∂x∂y\displaystyle \frac{\partial^{2}z}{\partial x\partial y} ∂z∂x=f1+yf2+2xf3∂2z∂x∂y=f11+xf12+f2+yf21+xyf22+2xf31+2x2f32=f2+f11+(x+y)f12+xyf22+2xf31+2x2f32 \begin{array}{ll} & \displaystyle \frac{\partial{z}}{\partial{x}} = f_{1} + yf_{2} + 2xf_{3}\\ & \displaystyle \frac{\partial^{2}{z}}{\partial{x}\partial{y}} = f_{11} + xf_{12} + f_{2} + yf_{21} + xyf_{22} + 2xf_{31} + 2x^{2}f_{32} \\ & = f_{2} + f_{11} + (x+y)f_{12}+ xyf_{22} + 2xf_{31}+ 2x^{2}f_{32} \end{array}

隐函数

例

  1. F(x,y)=0F(x,y) = 0 ==> 一个一元
  2. F(x,y,z)=0F(x,y,z)=0 ==> 一个二元
  3. {F(x,y,z)=0G(x,y,z)=0\begin{cases}F(x,y,z)=0\\G(x,y,z)=0\end{cases} ==> 二个一元 {y=y(x)z=z(x)\begin{cases}y=y(x)\\z=z(x)\end{cases}
  4. {F(x,y,u,v)=0G(x,y,u,v)=0\begin{cases}F(x,y,u,v) = 0\\ G(x,y,u,v)=0\end{cases} ==> 二个二元 {u=u(x,y)v=v(x,y)\begin{cases}u = u(x,y)\\v=v(x,y)\end{cases}

  1. {x+2y−z=3x2+4y2−z2=21\begin{cases}x+2y - z =3 \\ x^{2} + 4y^{2}-z^{2} = 21\end{cases} 求 dzdx\displaystyle \frac{dz}{dx} {x+2y−z=3x2+4y2−z2=21⇒{y=y(x)z=z(x){1+2dydx−dzdx=02x+8ydydx−2zdzdx=0 \begin{array}{ll} & \displaystyle \begin{cases} x + 2y -z =3 \\ x^{2} + 4y^{2} - z^{2} = 21 \end{cases} \Rightarrow \begin{cases} y = y(x) \\ z = z(x) \end{cases}\\\\ & \begin{cases} \displaystyle 1 + 2 \frac{dy}{dx} - \frac{dz}{dx} = 0 \\ \displaystyle 2x + 8y \frac{dy}{dx} - 2z \frac{dz}{dx} = 0 \end{cases} \\ \end{array}

  2. y=f(x,t),F(x2,sin⁡y,et)=0\displaystyle y = f(x,t),F(x^{2}, \sin y, e^{t}) = 0，f,Ff,F 连续可偏导，求 dtdx\displaystyle \frac{dt}{dx} {y−f(x,t)=0F(x2,sin⁡y,et)=0⇒{y=y(x)t=t(x){dydx−f1−dtdxf2=02xF1+F2cos⁡ydydx+F3etdtdx=0 \begin{array}{ll} & \begin{cases} y-f(x,t) = 0 \\ F(x^{2}, \sin y, e^{t}) = 0 \end{cases} \Rightarrow \begin{cases} y = y(x) \\ t = t(x) \end{cases} \\\\ & \begin{cases} \frac{dy}{dx} - f_{1} - \frac{dt}{dx}f_{2} = 0 \\ 2xF_{1} + F_{2}\cos y \frac{dy}{dx} + F_{3} e^{t}\frac{dt}{dx} = 0 \end{cases} \end{array}

  3. {xu−yv=1u2+yv=x2+yu\begin{cases}xu-yv = 1\\u^{2} +yv = x^{2} + yu\end{cases} 求 ∂u∂x,∂v∂x\displaystyle \frac{\partial u}{\partial x}, \frac{\partial v}{\partial x} {xu−yv−1=0u2+yv−x2−yu=0⇒{u=u(x,y)v=v(x,y){u+x∂u∂x−y∂v∂x=02u∂u∂x+y∂v∂x−2x−y∂u∂x=0 \begin{array}{ll} & \begin{cases}xu-yv - 1 = 0 \\u^{2} +yv - x^{2} - yu = 0\end{cases} \Rightarrow \begin{cases} u = u(x,y) \\ v = v(x,y) \end{cases} \\\\ & \begin{cases} u+x \frac{\partial{u}}{\partial{x}} - y \frac{\partial{v}}{\partial{x}} =0\\ 2u \frac{\partial{u}}{\partial{x}} + y \frac{\partial{v}}{\partial{x}} - 2x - y \frac{\partial{u}}{\partial{x}} = 0 \end{cases} \end{array}

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