文章目录

  • 一、多元向量值函数导数与微分的定义
  • 二、多元向量值函数的微分运算法则
  • 三、由方程所确定的隐函数的微分法

一、多元向量值函数导数与微分的定义

设有nnn元向量值函数f:U(x0)⊆Rn→Rm\bm f:U(x_0)\subseteq\mathbb{R}^n\to\mathbb{R}^mf:U(x0​)⊆Rn→Rm,该函数可以表示成
f(x)=[f1(x)f2(x)⋯fm(x)]=[f1(x1,x2,x3,⋯,xn)f2(x1,x2,x3,⋯,xn)⋯fm(x1,x2,x3,⋯,xn)]\bm f(\bm x)=\begin{bmatrix}f_1(\bm x)\\f_2(\bm x)\\\cdots\\f_m(\bm x)\end{bmatrix}=\begin{bmatrix}f_1(x_1,x_2,x_3,\cdots,x_n)\\f_2(x_1,x_2,x_3,\cdots,x_n)\\\cdots\\f_m(x_1,x_2,x_3,\cdots,x_n)\end{bmatrix}f(x)=⎣⎢⎢⎡​f1​(x)f2​(x)⋯fm​(x)​⎦⎥⎥⎤​=⎣⎢⎢⎡​f1​(x1​,x2​,x3​,⋯,xn​)f2​(x1​,x2​,x3​,⋯,xn​)⋯fm​(x1​,x2​,x3​,⋯,xn​)​⎦⎥⎥⎤​它将nnn维空间中的点x\bm xx映射为mmm维空间中的点f(x)\bm f(\bm x)f(x)。若f\bm ff的每个分量f1,f2,⋯,fnf_1,f_2,\cdots,f_nf1​,f2​,⋯,fn​都在点x0\bm x_0x0​处可微,则我们定义f\bm ff在x0\bm x_0x0​处的导数雅可比矩阵)为Df(x0)=[∂f1(x0)∂x1∂f1(x0)∂x2⋯∂f1(x0)∂xn∂f2(x0)∂x1∂f2(x0)∂x2⋯∂f2(x0)∂xn⋮⋮⋮∂fm(x0)∂x1∂fm(x0)∂x2⋯∂fm(x0)∂xn]=[∇f1(x0)∇f2(x0)⋮∇fm(x0)]\rm D\bm f(\bm x_0)=\begin{bmatrix}\frac{\partial f_1(\bm x_0)}{\partial x_1}&\frac{\partial f_1(\bm x_0)}{\partial x_2}&\cdots&\frac{\partial f_1(\bm x_0)}{\partial x_n}\\\frac{\partial f_2(\bm x_0)}{\partial x_1}&\frac{\partial f_2(\bm x_0)}{\partial x_2}&\cdots&\frac{\partial f_2(\bm x_0)}{\partial x_n}\\\vdots&\vdots&&\vdots\\\frac{\partial f_m(\bm x_0)}{\partial x_1}&\frac{\partial f_m(\bm x_0)}{\partial x_2}&\cdots&\frac{\partial f_m(\bm x_0)}{\partial x_n}\end{bmatrix}=\begin{bmatrix}\nabla f_1(\bm x_0)\\\nabla f_2(\bm x_0)\\\vdots\\\nabla f_m(\bm x_0)\end{bmatrix}Df(x0​)=⎣⎢⎢⎢⎢⎡​∂x1​∂f1​(x0​)​∂x1​∂f2​(x0​)​⋮∂x1​∂fm​(x0​)​​∂x2​∂f1​(x0​)​∂x2​∂f2​(x0​)​⋮∂x2​∂fm​(x0​)​​⋯⋯⋯​∂xn​∂f1​(x0​)​∂xn​∂f2​(x0​)​⋮∂xn​∂fm​(x0​)​​⎦⎥⎥⎥⎥⎤​=⎣⎢⎢⎢⎡​∇f1​(x0​)∇f2​(x0​)⋮∇fm​(x0​)​⎦⎥⎥⎥⎤​定义f\bm ff在x0\bm x_0x0​处的微分为df(x0)=[df1(x0)df2(x0)⋮dfm(x0)]=[∂f1(x0)∂x1dx1+∂f1(x0)∂x2dx2+⋯+∂f1(x0)∂xndxn∂f2(x0)∂x1dx1+∂f2(x0)∂x2dx2+⋯+∂f2(x0)∂xndxn⋮∂fm(x0)∂x1dx1+∂fm(x0)∂x2dx2+⋯+∂fm(x0)∂xndxn]=[∂f1(x0)∂x1∂f1(x0)∂x2⋯∂f1(x0)∂xn∂f2(x0)∂x1∂f2(x0)∂x2⋯∂f2(x0)∂xn⋮⋮⋮∂fm(x0)∂x1∂fm(x0)∂x2⋯∂fm(x0)∂xn][dx1dx2⋯dxn]=Df(x0)dx\begin{aligned}\text{d}\bm f(\bm x_0)&=\begin{bmatrix}\text{d}f_1(\bm x_0)\\\text{d}f_2(\bm x_0)\\\vdots\\\text{d}f_m(\bm x_0)\end{bmatrix}\\&=\begin{bmatrix}\frac{\partial f_1(\bm x_0)}{\partial x_1}\text{d}x_1+\frac{\partial f_1(\bm x_0)}{\partial x_2}\text{d}x_2+\cdots+\frac{\partial f_1(\bm x_0)}{\partial x_n}\text{d}x_n\\\frac{\partial f_2(\bm x_0)}{\partial x_1}\text{d}x_1+\frac{\partial f_2(\bm x_0)}{\partial x_2}\text{d}x_2+\cdots+\frac{\partial f_2(\bm x_0)}{\partial x_n}\text{d}x_n\\\vdots\\\frac{\partial f_m(\bm x_0)}{\partial x_1}\text{d}x_1+\frac{\partial f_m(\bm x_0)}{\partial x_2}\text{d}x_2+\cdots+\frac{\partial f_m(\bm x_0)}{\partial x_n}\text{d}x_n\end{bmatrix}\\&=\begin{bmatrix}\frac{\partial f_1(\bm x_0)}{\partial x_1}&\frac{\partial f_1(\bm x_0)}{\partial x_2}&\cdots&\frac{\partial f_1(\bm x_0)}{\partial x_n}\\\frac{\partial f_2(\bm x_0)}{\partial x_1}&\frac{\partial f_2(\bm x_0)}{\partial x_2}&\cdots&\frac{\partial f_2(\bm x_0)}{\partial x_n}\\\vdots&\vdots&&\vdots\\\frac{\partial f_m(\bm x_0)}{\partial x_1}&\frac{\partial f_m(\bm x_0)}{\partial x_2}&\cdots&\frac{\partial f_m(\bm x_0)}{\partial x_n}\end{bmatrix}\begin{bmatrix}\text{d}x_1\\\text{d}x_2\\\cdots\\\text{d}x_n\end{bmatrix}\\&=\text{D}\bm{f}(\bm x_0)\text{d}\bm{x}\end{aligned}df(x0​)​=⎣⎢⎢⎢⎡​df1​(x0​)df2​(x0​)⋮dfm​(x0​)​⎦⎥⎥⎥⎤​=⎣⎢⎢⎢⎢⎡​∂x1​∂f1​(x0​)​dx1​+∂x2​∂f1​(x0​)​dx2​+⋯+∂xn​∂f1​(x0​)​dxn​∂x1​∂f2​(x0​)​dx1​+∂x2​∂f2​(x0​)​dx2​+⋯+∂xn​∂f2​(x0​)​dxn​⋮∂x1​∂fm​(x0​)​dx1​+∂x2​∂fm​(x0​)​dx2​+⋯+∂xn​∂fm​(x0​)​dxn​​⎦⎥⎥⎥⎥⎤​=⎣⎢⎢⎢⎢⎡​∂x1​∂f1​(x0​)​∂x1​∂f2​(x0​)​⋮∂x1​∂fm​(x0​)​​∂x2​∂f1​(x0​)​∂x2​∂f2​(x0​)​⋮∂x2​∂fm​(x0​)​​⋯⋯⋯​∂xn​∂f1​(x0​)​∂xn​∂f2​(x0​)​⋮∂xn​∂fm​(x0​)​​⎦⎥⎥⎥⎥⎤​⎣⎢⎢⎡​dx1​dx2​⋯dxn​​⎦⎥⎥⎤​=Df(x0​)dx​当m=nm=nm=n时,雅可比矩阵为方阵,该方阵的行列式称为f\bm ff在x0\bm x_0x0​处的雅可比行列式(Jacobian Determinant,简称雅可比式),记作Jf(x0)=∣Df(x0)∣=∂(f1,f2,⋯,fn)∂(x1,x2,⋯,xn)∣x0\bm J_f(\bm x_0)=\left|\rm D\bm f(\bm x_0)\right|=\left.\frac{\partial(f_1,f_2,\cdots,f_n)}{\partial(x_1,x_2,\cdots,x_n)}\right|_{\bm x_0}Jf​(x0​)=∣Df(x0​)∣=∂(x1​,x2​,⋯,xn​)∂(f1​,f2​,⋯,fn​)​∣∣∣∣​x0​​例如,设{x(ρ,θ)=ρcos⁡θy(ρ,θ)=ρsin⁡θ\begin{cases}x(\rho,\theta)=\rho\cos\theta\\y(\rho,\theta)=\rho\sin\theta\end{cases}{x(ρ,θ)=ρcosθy(ρ,θ)=ρsinθ​则∂(x,y)∂(ρ,θ)=∣cos⁡θ−ρsin⁡θsin⁡θρcos⁡θ∣=ρ(cos⁡2θ+sin⁡2θ)=ρ\frac{\partial(x,y)}{\partial(\rho,\theta)}=\begin{vmatrix}\cos\theta&-\rho\sin\theta\\\sin\theta&\rho\cos\theta\end{vmatrix}=\rho(\cos^2\theta+\sin^2\theta)=\rho∂(ρ,θ)∂(x,y)​=∣∣∣∣​cosθsinθ​−ρsinθρcosθ​∣∣∣∣​=ρ(cos2θ+sin2θ)=ρ设x0=(x1,x2,⋯,xn)T\bm x_0=(x_1,x_2,\cdots,x_n)^Tx0​=(x1​,x2​,⋯,xn​)T,我们定义f\bm ff在x0\bm x_0x0​处对xix_ixi​的偏导数为∂f(x0)∂xi=lim⁡Δxi→0f([x1x2⋮xi+Δxi⋮xn])−f([x1x2⋮xi⋮xn])Δxi\frac{\partial\bm f(\bm x_0)}{\partial x_i}=\lim\limits_{\Delta x_i\to0}\frac{\bm f\left(\begin{bmatrix}x_1\\x_2\\\vdots\\x_i+\Delta x_i\\\vdots\\x_n\end{bmatrix}\right)-\bm f\left(\begin{bmatrix}x_1\\x_2\\\vdots\\x_i\\\vdots\\x_n\end{bmatrix}\right)}{\Delta x_i}∂xi​∂f(x0​)​=Δxi​→0lim​Δxi​f⎝⎜⎜⎜⎜⎜⎜⎜⎜⎛​⎣⎢⎢⎢⎢⎢⎢⎢⎢⎡​x1​x2​⋮xi​+Δxi​⋮xn​​⎦⎥⎥⎥⎥⎥⎥⎥⎥⎤​⎠⎟⎟⎟⎟⎟⎟⎟⎟⎞​−f⎝⎜⎜⎜⎜⎜⎜⎜⎜⎛​⎣⎢⎢⎢⎢⎢⎢⎢⎢⎡​x1​x2​⋮xi​⋮xn​​⎦⎥⎥⎥⎥⎥⎥⎥⎥⎤​⎠⎟⎟⎟⎟⎟⎟⎟⎟⎞​​当∂f(x0)∂xi\frac{\partial\bm f(\bm x_0)}{\partial x_i}∂xi​∂f(x0​)​存在时,有∂f(x0)∂xi=[∂f1(x0)∂xi∂f2(x0)∂xi⋮∂fm(x0)∂xi]\frac{\partial\bm f(\bm x_0)}{\partial x_i}=\begin{bmatrix}\frac{\partial f_1(\bm x_0)}{\partial x_i}\\\frac{\partial f_2(\bm x_0)}{\partial x_i}\\\vdots\\\frac{\partial f_m(\bm x_0)}{\partial x_i}\end{bmatrix}∂xi​∂f(x0​)​=⎣⎢⎢⎢⎢⎡​∂xi​∂f1​(x0​)​∂xi​∂f2​(x0​)​⋮∂xi​∂fm​(x0​)​​⎦⎥⎥⎥⎥⎤​因此我们可以把雅可比矩阵Df(x0)\rm D\bm f(\bm x_0)Df(x0​)按列和按行分块分别写作Df(x0)=[∂f(x0)∂x1∂f(x0)∂x2⋯∂f(x0)∂xn]=[∇f1(x0)∇f2(x0)⋮∇fm(x0)]\rm D\bm f(\bm x_0)=\begin{bmatrix}\frac{\partial\bm f(\bm x_0)}{\partial x_1}&\frac{\partial\bm f(\bm x_0)}{\partial x_2}&\cdots&\frac{\partial\bm f(\bm x_0)}{\partial x_n}\end{bmatrix}=\begin{bmatrix}\nabla f_1(\bm x_0)\\\nabla f_2(\bm x_0)\\\vdots\\\nabla f_m(\bm x_0)\end{bmatrix}Df(x0​)=[∂x1​∂f(x0​)​​∂x2​∂f(x0​)​​⋯​∂xn​∂f(x0​)​​]=⎣⎢⎢⎢⎡​∇f1​(x0​)∇f2​(x0​)⋮∇fm​(x0​)​⎦⎥⎥⎥⎤​

二、多元向量值函数的微分运算法则

定理1 设向量值函数f\bm ff和g\bm gg都在点x\bm xx处可微,uuu是在x\bm xx处可微的数量值函数,则
(1) f+g\bm f+\bm gf+g在xxx处可微,且其导数为D(f+g)(x)=Df(x)+Dg(x)\text{D}(\bm f+\bm g)(\bm x)=\rm D\bm f(\bm x)+\rm D\bm g(\bm x)D(f+g)(x)=Df(x)+Dg(x)(2) ⟨f,g⟩\left\langle\bm f,\bm g\right\rangle⟨f,g⟩在x\bm xx处可微,且其导数为D⟨f,g⟩(x)=(f(x))TDg(x)+(g(x))TDf(x)\rm D\left\langle\bm f,\bm g\right\rangle(\bm x)=(\bm f(\bm x))^T\rm D\bm g(\bm x)+(\bm g(\bm x))^T\rm D\bm f(\bm x)D⟨f,g⟩(x)=(f(x))TDg(x)+(g(x))TDf(x)(3) ufu\bm fuf在x\bm xx处可微,且其导数为D(uf)(x)=u(x)Df(x)+f(x)∇u(x)\text{D}(u\bm f)(\bm x)=u(\bm x)\text D\bm f(\bm x)+\bm f(\bm x)\nabla u(\bm x)D(uf)(x)=u(x)Df(x)+f(x)∇u(x)(4) 若f:R→R3,g:R→R3\bm f:\mathbb{R}\to\mathbb{R}^3,\bm g:\mathbb{R}\to\mathbb{R}^3f:R→R3,g:R→R3,则向量积f×g\bm f\times\bm gf×g在xxx处可微,且其导数为D(f×g)(x)=Df(x)×g(x)+f(x)×Dg(x)\text{D}(\bm f\times\bm g)(x)=\text{D}\bm f(x)\times\bm g(x)+\bm f(x)\times\text{D}\bm g(x)D(f×g)(x)=Df(x)×g(x)+f(x)×Dg(x)证明:
(1) 显然成立。
(2) 设f=[f1f2⋮fm],g=[g1g2⋮gm]\bm f=\begin{bmatrix}f_1\\f_2\\\vdots\\f_m\end{bmatrix},\bm g=\begin{bmatrix}g_1\\g_2\\\vdots\\g_m\end{bmatrix}f=⎣⎢⎢⎢⎡​f1​f2​⋮fm​​⎦⎥⎥⎥⎤​,g=⎣⎢⎢⎢⎡​g1​g2​⋮gm​​⎦⎥⎥⎥⎤​,则数量值函数F=⟨f,g⟩(x)=∑i=1mfigiF=\left\langle\bm f,\bm g\right\rangle(\bm x)=\sum\limits_{i=1}^mf_ig_iF=⟨f,g⟩(x)=i=1∑m​fi​gi​,且DF(x)=∇F(x)=∑i=1m∇figi(x)\begin{aligned}\text{D}F(\bm x)&=\nabla F(\bm x)\\&=\sum\limits_{i=1}^m\nabla f_ig_i(\bm x)\end{aligned}DF(x)​=∇F(x)=i=1∑m​∇fi​gi​(x)​我们知道,对于j∈{1,2,⋯,n}j\in\{1,2,\cdots,n\}j∈{1,2,⋯,n},∂figi(xj)∂xj=fi∂gi(xj)∂xj+gi∂fi(xj)∂xj\frac{\partial f_ig_i(x_j)}{\partial x_j}=f_i\frac{\partial g_i(x_j)}{\partial x_j}+g_i\frac{\partial f_i(x_j)}{\partial x_j}∂xj​∂fi​gi​(xj​)​=fi​∂xj​∂gi​(xj​)​+gi​∂xj​∂fi​(xj​)​故∇figi(x)=(fi∂gi(x1)∂x1+gi∂fi(x1)∂x1,fi∂gi(x2)∂x2+gi∂fi(x2)∂x2,⋯,fi∂gi(xn)∂xn+gi∂fi(xn)∂xn)=fi(x)∇gi(x)+∇fi(x)gi(x)\begin{aligned}\nabla f_ig_i(\bm x)&=\left(f_i\frac{\partial g_i(x_1)}{\partial x_1}+g_i\frac{\partial f_i(x_1)}{\partial x_1},f_i\frac{\partial g_i(x_2)}{\partial x_2}+g_i\frac{\partial f_i(x_2)}{\partial x_2},\cdots,f_i\frac{\partial g_i(x_n)}{\partial x_n}+g_i\frac{\partial f_i(x_n)}{\partial x_n}\right)\\&=f_i(\bm x)\nabla g_i(\bm x)+\nabla f_i(\bm x)g_i(\bm x)\end{aligned}∇fi​gi​(x)​=(fi​∂x1​∂gi​(x1​)​+gi​∂x1​∂fi​(x1​)​,fi​∂x2​∂gi​(x2​)​+gi​∂x2​∂fi​(x2​)​,⋯,fi​∂xn​∂gi​(xn​)​+gi​∂xn​∂fi​(xn​)​)=fi​(x)∇gi​(x)+∇fi​(x)gi​(x)​那么DF(x)=∑i=1m[fi(x)∇gi(x)+∇fi(x)gi(x)]=[f1(x)f2(x)⋯fm(x)][∇g1(x)∇g2(x)⋮∇gm(x)]+[g1(x)g2(x)⋯gm(x)][∇f1(x)∇f2(x)⋮∇fm(x)]=(f(x))TDg(x)+(g(x))TDf(x)\begin{aligned}\text{D}F(\bm x)&=\sum\limits_{i=1}^m[f_i(\bm x)\nabla g_i(\bm x)+\nabla f_i(\bm x)g_i(\bm x)]\\&=\begin{bmatrix}f_1(\bm x)&f_2(\bm x)&\cdots&f_m(\bm x)\end{bmatrix}\begin{bmatrix}\nabla g_1(\bm x)\\\nabla g_2(\bm x)\\\vdots\\\nabla g_m(\bm x)\end{bmatrix}+\begin{bmatrix}g_1(\bm x)&g_2(\bm x)&\cdots&g_m(\bm x)\end{bmatrix}\begin{bmatrix}\nabla f_1(\bm x)\\\nabla f_2(\bm x)\\\vdots\\\nabla f_m(\bm x)\end{bmatrix}\\&=(\bm f(\bm x))^T\rm D\bm g(\bm x)+(\bm g(\bm x))^T\rm D\bm f(\bm x)\end{aligned}DF(x)​=i=1∑m​[fi​(x)∇gi​(x)+∇fi​(x)gi​(x)]=[f1​(x)​f2​(x)​⋯​fm​(x)​]⎣⎢⎢⎢⎡​∇g1​(x)∇g2​(x)⋮∇gm​(x)​⎦⎥⎥⎥⎤​+[g1​(x)​g2​(x)​⋯​gm​(x)​]⎣⎢⎢⎢⎡​∇f1​(x)∇f2​(x)⋮∇fm​(x)​⎦⎥⎥⎥⎤​=(f(x))TDg(x)+(g(x))TDf(x)​
(3) 是(2)的特例。
(4) D(f×g)(x)=D∣ijkf1f2f3g1g2g3∣=ddx[f2g3−f3g2f3g1−f1g3f1g2−f2g1]=[f2′g3+f2g3′−f3′g2−f3g2′f3′g1+f3g1′−f1′g3−f1g3′f1′g2+f1g2′−f2′g1−f2g1′]=[f2′g3−f3′g2f3′g1−f1′g3f1′g2−f2′g1]+[f2g3′−f3g2′f3g1′−f1g3′f1g2′−f2g1′]=[f1′f2′f3′]×[g1g2g3]+[f1f2f3]×[g1′g2′g3′]=Df(x)×g(x)+f(x)×Dg(x)\begin{aligned}\text{D}(\bm f\times\bm g)(x)&=\text{D}\begin{vmatrix}\bm i&\bm j&\bm k\\f_1&f_2&f_3\\g_1&g_2&g_3\end{vmatrix}\\&=\frac{\rm d}{\text{d}x}\begin{bmatrix}f_2g_3-f_3g_2\\f_3g_1-f_1g_3\\f_1g_2-f_2g_1\end{bmatrix}\\&=\begin{bmatrix}f_2'g_3+f_2g_3'-f_3'g_2-f_3g_2'\\f_3'g_1+f_3g_1'-f_1'g_3-f_1g_3'\\f_1'g_2+f_1g_2'-f_2'g_1-f_2g_1'\end{bmatrix}\\&=\begin{bmatrix}f_2'g_3-f_3'g_2\\f_3'g_1-f_1'g_3\\f_1'g_2-f_2'g_1\end{bmatrix}+\begin{bmatrix}f_2g_3'-f_3g_2'\\f_3g_1'-f_1g_3'\\f_1g_2'-f_2g_1'\end{bmatrix}\\&=\begin{bmatrix}f_1'\\f_2'\\f_3'\end{bmatrix}\times\begin{bmatrix}g_1\\g_2\\g_3\end{bmatrix}+\begin{bmatrix}f_1\\f_2\\f_3\end{bmatrix}\times\begin{bmatrix}g_1'\\g_2'\\g_3'\end{bmatrix}\\&=\text{D}\bm f(x)\times\bm g(x)+\bm f(x)\times\text{D}\bm g(x)\end{aligned}D(f×g)(x)​=D∣∣∣∣∣∣​if1​g1​​jf2​g2​​kf3​g3​​∣∣∣∣∣∣​=dxd​⎣⎡​f2​g3​−f3​g2​f3​g1​−f1​g3​f1​g2​−f2​g1​​⎦⎤​=⎣⎡​f2′​g3​+f2​g3′​−f3′​g2​−f3​g2′​f3′​g1​+f3​g1′​−f1′​g3​−f1​g3′​f1′​g2​+f1​g2′​−f2′​g1​−f2​g1′​​⎦⎤​=⎣⎡​f2′​g3​−f3′​g2​f3′​g1​−f1′​g3​f1′​g2​−f2′​g1​​⎦⎤​+⎣⎡​f2​g3′​−f3​g2′​f3​g1′​−f1​g3′​f1​g2′​−f2​g1′​​⎦⎤​=⎣⎡​f1′​f2′​f3′​​⎦⎤​×⎣⎡​g1​g2​g3​​⎦⎤​+⎣⎡​f1​f2​f3​​⎦⎤​×⎣⎡​g1′​g2′​g3′​​⎦⎤​=Df(x)×g(x)+f(x)×Dg(x)​证毕。∎

定理2 设r=r(t)\bm r=\bm r(t)r=r(t)表示空间中动点[x(t)y(t)z(t)]\begin{bmatrix}x(t)\\y(t)\\z(t)\end{bmatrix}⎣⎡​x(t)y(t)z(t)​⎦⎤​的向径,则⟨r′(t),r(t)⟩≡0⟺∥r(t)∥≡c\left\langle\bm r'(t),\bm r(t)\right\rangle\equiv0\Longleftrightarrow\|\bm r(t)\|\equiv c⟨r′(t),r(t)⟩≡0⟺∥r(t)∥≡c其中ccc为常数,这表示动点的轨迹在以原点为中心的球面上。
证明
由定理1(2)知ddt⟨r,r⟩(t)=⟨r,r′⟩(t)+⟨r′,r⟩(t)\frac{\rm d}{\text{d}t}\left\langle\bm r,\bm r\right\rangle(t)=\left\langle\bm r,\bm r'\right\rangle(t)+\left\langle\bm r',\bm r\right\rangle(t)dtd​⟨r,r⟩(t)=⟨r,r′⟩(t)+⟨r′,r⟩(t)即ddt∥r(t)∥2=2⟨r′(t),r(t)⟩\frac{\rm d}{\text{d}t}\|\bm r(t)\|^2=2\left\langle\bm r'(t),\bm r(t)\right\rangledtd​∥r(t)∥2=2⟨r′(t),r(t)⟩故⟨r′(t),r(t)⟩≡0⟺ddt∥r(t)∥2≡0⟺∥r(t)∥2≡C⟺∥r(t)∥≡c\left\langle\bm r'(t),\bm r(t)\right\rangle\equiv0\Longleftrightarrow\frac{\rm d}{\text{d}t}\|\bm r(t)\|^2\equiv0\Longleftrightarrow\|\bm r(t)\|^2\equiv C\Longleftrightarrow\|\bm r(t)\|\equiv c⟨r′(t),r(t)⟩≡0⟺dtd​∥r(t)∥2≡0⟺∥r(t)∥2≡C⟺∥r(t)∥≡c。∎

定理3(向量值函数的链式法则) 设向量值函数g(x):Rn→Rp\bm g(\bm x):\mathbb{R}^n\to\mathbb{R}^pg(x):Rn→Rp在点x0\bm x_0x0​处可微,向量值函数f(u):Rp→Rm\bm f(\bm u):\mathbb{R}^p\to\mathbb{R}^mf(u):Rp→Rm在点g(x0)\bm g(\bm x_0)g(x0​)处可微,则复合函数w(x)=f(g(x))\bm w(\bm x)=\bm f(\bm g(\bm x))w(x)=f(g(x))在点x0\bm x_0x0​处可微,且Dw(x0)=Df(g(x0))Dg(x0)\rm D\bm w(\bm x_0)=\rm D\bm f(\bm g(\bm x_0))\rm D\bm g(\bm x_0)Dw(x0​)=Df(g(x0​))Dg(x0​)证明提要:由复合函数的求导法则,Dw(x)\rm D\bm w(\bm x)Dw(x)的第(i,j)(i,j)(i,j)个元素∂wi∂xj\frac{\partial w_i}{\partial x_j}∂xj​∂wi​​可以表示为∂wi∂xj=∂fi(g(x))∂xj=∂fi(g1(x1,x2,⋯,xn),g2(x1,x2,⋯,xn),⋯,gp(x1,x2,⋯,xn))∂xj=∑k=1p∂fi∂uk∂gk∂xj\begin{aligned}\frac{\partial w_i}{\partial x_j}&=\frac{\partial f_i(\bm g(\bm x))}{\partial x_j}\\&=\frac{\partial f_i(g_1(x_1,x_2,\cdots,x_n),g_2(x_1,x_2,\cdots,x_n),\cdots,g_p(x_1,x_2,\cdots,x_n))}{\partial x_j}\\&=\sum\limits_{k=1}^{p}\frac{\partial f_i}{\partial u_k}\frac{\partial g_k}{\partial x_j}\end{aligned}∂xj​∂wi​​​=∂xj​∂fi​(g(x))​=∂xj​∂fi​(g1​(x1​,x2​,⋯,xn​),g2​(x1​,x2​,⋯,xn​),⋯,gp​(x1​,x2​,⋯,xn​))​=k=1∑p​∂uk​∂fi​​∂xj​∂gk​​​恰好是矩阵乘法的形式。∎

当n=p=mn=p=mn=p=m时,取两端的行列式得∂(w1,w2,⋯,wn)∂(x1,x2,⋯,xn)=∂(w1,w2,⋯,wn)∂(u1,u2,⋯,un)∂(g1,g2,⋯,gn)∂(x1,x2,⋯,xn)\frac{\partial(w_1,w_2,\cdots,w_n)}{\partial(x_1,x_2,\cdots,x_n)}=\frac{\partial(w_1,w_2,\cdots,w_n)}{\partial(u_1,u_2,\cdots,u_n)}\frac{\partial(g_1,g_2,\cdots,g_n)}{\partial(x_1,x_2,\cdots,x_n)}∂(x1​,x2​,⋯,xn​)∂(w1​,w2​,⋯,wn​)​=∂(u1​,u2​,⋯,un​)∂(w1​,w2​,⋯,wn​)​∂(x1​,x2​,⋯,xn​)∂(g1​,g2​,⋯,gn​)​即复合函数的雅可比式是两个函数的雅可比式的乘积(Jw=JfJg\bm J_w=\bm J_f\bm J_gJw​=Jf​Jg​)。

推论 设f,g:Rn→Rn\bm f,\bm g:\mathbb{R}^n\to\mathbb{R}^nf,g:Rn→Rn互为反函数,则它们的雅可比式互为倒数。
证明:令w(x)=f(g(x))≡x\bm w(\bm x)=\bm f(\bm g(\bm x))\equiv \bm xw(x)=f(g(x))≡x,而Dw(x)=I\rm D\bm w(\bm x)=\bm IDw(x)=I,故Jw=1\bm J_w=1Jw​=1,JfJg=1\bm J_f\bm J_g=1Jf​Jg​=1。∎

三、由方程所确定的隐函数的微分法

例1 已知方程组{F1(x,y,u,v)=xu+yv=0F2(x,y,u,v)=yu+xv=1\begin{cases}F_1(x,y,u,v)=xu+yv=0\\F_2(x,y,u,v)=yu+xv=1\end{cases}{F1​(x,y,u,v)=xu+yv=0F2​(x,y,u,v)=yu+xv=1​确定了隐函数{u=u(x,y)v=v(x,y)\begin{cases}u=u(x,y)\\v=v(x,y)\end{cases}{u=u(x,y)v=v(x,y)​,求(1)∂u∂x\partial u\over\partial x∂x∂u​及(2)∂v∂y\partial v\over\partial y∂y∂v​。

(1) 方程组两边对xxx求偏导得{u+x∂u∂x+y∂v∂x=0y∂u∂x+v+x∂v∂x=0\begin{cases}u+x{\partial u\over\partial x}+y{\partial v\over\partial x}=0\\y{\partial u\over\partial x}+v+x{\partial v\over\partial x}=0\end{cases}{u+x∂x∂u​+y∂x∂v​=0y∂x∂u​+v+x∂x∂v​=0​这是一个关于∂u∂x,∂v∂x{\partial u\over\partial x},{\partial v\over\partial x}∂x∂u​,∂x∂v​的二元一次方程组,可以表示成[xyyx][∂u∂x∂v∂x]=[−u−v]\begin{bmatrix}x&y\\y&x\end{bmatrix}\begin{bmatrix}{\partial u\over\partial x}\\{\partial v\over\partial x}\end{bmatrix}=\begin{bmatrix}-u\\-v\end{bmatrix}[xy​yx​][∂x∂u​∂x∂v​​]=[−u−v​]其中系数行列式即为雅可比式J=∂(F1,F2)∂(u,v)\bm J=\frac{\partial(F_1,F_2)}{\partial(u,v)}J=∂(u,v)∂(F1​,F2​)​。根据Cramer法则,∂u∂x=JuJ=∣−uy−vx∣∣xyyx∣=vy−uxx2−y2\frac{\partial u}{\partial x}=\frac{\bm J_u}{\bm J}=\frac{\begin{vmatrix}-u&y\\-v&x\end{vmatrix}}{\begin{vmatrix}x&y\\y&x\end{vmatrix}}=\frac{vy-ux}{x^2-y^2}∂x∂u​=JJu​​=∣∣∣∣​xy​yx​∣∣∣∣​∣∣∣∣​−u−v​yx​∣∣∣∣​​=x2−y2vy−ux​(2) 同理可得∂v∂y=vy−uxx2−y2\frac{\partial v}{\partial y}=\frac{vy-ux}{x^2-y^2}∂y∂v​=x2−y2vy−ux​。注意,原方程中将uuu与vvv互换、xxx与yyy互换所得的方程与原方程完全相同,所以∂u∂x=∂v∂y\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}∂x∂u​=∂y∂v​。∎

例2 已知{u+v+w=xuv+vw+wu=yuvw=z\begin{cases}u+v+w=x\\uv+vw+wu=y\\uvw=z\end{cases}⎩⎪⎨⎪⎧​u+v+w=xuv+vw+wu=yuvw=z​,求∂u∂x,∂u∂y,∂u∂z\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial u}{\partial z}∂x∂u​,∂y∂u​,∂z∂u​。
:两边取微分得{du+dv+dw=dx(v+w)du+(u+w)dv+(u+v)dw=dyvwdu+uwdv+uvdw=dz\begin{cases}\text{d}u+\text{d}v+\text{d}w&=\text{d}x\\(v+w)\text{d}u+(u+w)\text{d}v+(u+v)\text{d}w&=\text{d}y\\vw\text{d}u+uw\text{d}v+uv\text{d}w&=\text{d}z\end{cases}⎩⎪⎨⎪⎧​du+dv+dw(v+w)du+(u+w)dv+(u+v)dwvwdu+uwdv+uvdw​=dx=dy=dz​即[111v+wu+wu+vvwuwuv][dudvdw]=[dxdydz]\begin{bmatrix}1&1&1\\v+w&u+w&u+v\\vw&uw&uv\end{bmatrix}\begin{bmatrix}\text{d}u\\\text{d}v\\\text{d}w\end{bmatrix}=\begin{bmatrix}\text{d}x\\\text{d}y\\\text{d}z\end{bmatrix}⎣⎡​1v+wvw​1u+wuw​1u+vuv​⎦⎤​⎣⎡​dudvdw​⎦⎤​=⎣⎡​dxdydz​⎦⎤​雅可比行列式J=∣111v+wu+wu+vvwuwuv∣=u2(v−w)+v2(w−u)+w2(u−v)=(u−v)(v−w)(u−w)\begin{aligned}\bm J&=\begin{vmatrix}1&1&1\\v+w&u+w&u+v\\vw&uw&uv\end{vmatrix}\\&=u^2(v-w)+v^2(w-u)+w^2(u-v)\\&=(u-v)(v-w)(u-w)\end{aligned}J​=∣∣∣∣∣∣​1v+wvw​1u+wuw​1u+vuv​∣∣∣∣∣∣​=u2(v−w)+v2(w−u)+w2(u−v)=(u−v)(v−w)(u−w)​故du=JuJ=∣dx11dyu+wu+vdzuwuv∣J=u2(v−w)dx−u(v−w)dy+(v−w)dzJ=u2dx−udy+dz(u−v)(u−w)\begin{aligned}\text{d}u&=\frac{\bm J_u}{\bm J}\\&=\frac{\begin{vmatrix}\text{d}x&1&1\\\text{d}y&u+w&u+v\\\text{d}z&uw&uv\end{vmatrix}}{\bm J}\\&=\frac{u^2(v-w)\text{d}x-u(v-w)\text{d}y+(v-w)\text{d}z}{\bm J}\\&=\frac{u^2\text{d}x-u\text{d}y+\text{d}z}{(u-v)(u-w)}\end{aligned}du​=JJu​​=J∣∣∣∣∣∣​dxdydz​1u+wuw​1u+vuv​∣∣∣∣∣∣​​=Ju2(v−w)dx−u(v−w)dy+(v−w)dz​=(u−v)(u−w)u2dx−udy+dz​​因此{∂u∂x=u2(u−v)(u−w)∂u∂y=−u(u−v)(u−w)∂u∂z=1(u−v)(u−w)\begin{cases}\frac{\partial u}{\partial x}=\frac{u^2}{(u-v)(u-w)}\\\frac{\partial u}{\partial y}=-\frac{u}{(u-v)(u-w)}\\\frac{\partial u}{\partial z}=\frac{1}{(u-v)(u-w)}\end{cases}⎩⎪⎨⎪⎧​∂x∂u​=(u−v)(u−w)u2​∂y∂u​=−(u−v)(u−w)u​∂z∂u​=(u−v)(u−w)1​​∎

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