【高等数学笔记】多元向量值函数的导数与微分
文章目录一、多元向量值函数的导数与微分的定义二、多元向量值函数的微分运算法则一、多元向量值函数的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)f3(x)⋯fm(x)]=[f1(x1,x2,x3,⋯ ,xn)f
一、多元向量值函数导数与微分的定义
设有 n n n元向量值函数 f : U ( x 0 ) ⊆ R n → R m \bm f:U(x_0)\subseteq\mathbb{R}^n\to\mathbb{R}^m f:U(x0)⊆Rn→Rm，该函数可以表示成
f ( x ) = [ f 1 ( x ) f 2 ( x ) ⋯ f m ( x ) ] = [ f 1 ( x 1 , x 2 , x 3 , ⋯ , x n ) f 2 ( x 1 , x 2 , x 3 , ⋯ , x n ) ⋯ f m ( x 1 , x 2 , x 3 , ⋯ , x n ) ] \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 n n n维空间中的点 x \bm x x映射为 m m m维空间中的点 f ( x ) \bm f(\bm x) f(x)。若 f \bm f f的每个分量 f 1 , f 2 , ⋯ , f n f_1,f_2,\cdots,f_n f1,f2,⋯,fn都在点 x 0 \bm x_0 x0处可微，则我们定义 f \bm f f在 x 0 \bm x_0 x0处的导数（雅可比矩阵）为 D f ( x 0 ) = [ ∂ f 1 ( x 0 ) ∂ x 1 ∂ f 1 ( x 0 ) ∂ x 2 ⋯ ∂ f 1 ( x 0 ) ∂ x n ∂ f 2 ( x 0 ) ∂ x 1 ∂ f 2 ( x 0 ) ∂ x 2 ⋯ ∂ f 2 ( x 0 ) ∂ x n ⋮ ⋮ ⋮ ∂ f m ( x 0 ) ∂ x 1 ∂ f m ( x 0 ) ∂ x 2 ⋯ ∂ f m ( x 0 ) ∂ x n ] = [ ∇ f 1 ( x 0 ) ∇ f 2 ( x 0 ) ⋮ ∇ f m ( x 0 ) ] \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 f f在 x 0 \bm x_0 x0处的微分为 d f ( x 0 ) = [ d f 1 ( x 0 ) d f 2 ( x 0 ) ⋮ d f m ( x 0 ) ] = [ ∂ f 1 ( x 0 ) ∂ x 1 d x 1 + ∂ f 1 ( x 0 ) ∂ x 2 d x 2 + ⋯ + ∂ f 1 ( x 0 ) ∂ x n d x n ∂ f 2 ( x 0 ) ∂ x 1 d x 1 + ∂ f 2 ( x 0 ) ∂ x 2 d x 2 + ⋯ + ∂ f 2 ( x 0 ) ∂ x n d x n ⋮ ∂ f m ( x 0 ) ∂ x 1 d x 1 + ∂ f m ( x 0 ) ∂ x 2 d x 2 + ⋯ + ∂ f m ( x 0 ) ∂ x n d x n ] = [ ∂ f 1 ( x 0 ) ∂ x 1 ∂ f 1 ( x 0 ) ∂ x 2 ⋯ ∂ f 1 ( x 0 ) ∂ x n ∂ f 2 ( x 0 ) ∂ x 1 ∂ f 2 ( x 0 ) ∂ x 2 ⋯ ∂ f 2 ( x 0 ) ∂ x n ⋮ ⋮ ⋮ ∂ f m ( x 0 ) ∂ x 1 ∂ f m ( x 0 ) ∂ x 2 ⋯ ∂ f m ( x 0 ) ∂ x n ] [ d x 1 d x 2 ⋯ d x n ] = D f ( x 0 ) d x \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)
dx1dx2⋯dxn
=Df(x0)dx当 m = n m=n m=n时，雅可比矩阵为方阵，该方阵的行列式称为 f \bm f f在 x 0 \bm x_0 x0处的雅可比行列式（Jacobian Determinant，简称雅可比式），记作 J f ( x 0 ) = ∣ D f ( x 0 ) ∣ = ∂ ( f 1 , f 2 , ⋯ , f n ) ∂ ( x 1 , x 2 , ⋯ , x n ) ∣ x 0 \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 ) = ρ cos θ y ( ρ , θ ) = ρ sin 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 − ρ sin θ sin θ ρ cos θ ∣ = ρ ( cos 2 θ + sin 2 θ ) = 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 x 0 = ( x 1 , x 2 , ⋯ , x n ) T \bm x_0=(x_1,x_2,\cdots,x_n)^T x0=(x1,x2,⋯,xn)T，我们定义 f \bm f f在 x 0 \bm x_0 x0处对 x i x_i xi的偏导数为 ∂ f ( x 0 ) ∂ x i = lim Δ x i → 0 f ( [ x 1 x 2 ⋮ x i + Δ x i ⋮ x n ] ) − f ( [ x 1 x 2 ⋮ x i ⋮ x n ] ) Δ x i \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→0limΔxif
x1x2⋮xi+Δxi⋮xn
−f
x1x2⋮xi⋮xn
当 ∂ f ( x 0 ) ∂ x i \frac{\partial\bm f(\bm x_0)}{\partial x_i} ∂xi∂f(x0)存在时，有 ∂ f ( x 0 ) ∂ x i = [ ∂ f 1 ( x 0 ) ∂ x i ∂ f 2 ( x 0 ) ∂ x i ⋮ ∂ f m ( x 0 ) ∂ x i ] \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 D f ( x 0 ) \rm D\bm f(\bm x_0) Df(x0)按列和按行分块分别写作 D f ( x 0 ) = [ ∂ f ( x 0 ) ∂ x 1 ∂ f ( x 0 ) ∂ x 2 ⋯ ∂ f ( x 0 ) ∂ x n ] = [ ∇ f 1 ( x 0 ) ∇ f 2 ( x 0 ) ⋮ ∇ f m ( x 0 ) ] \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 f f和 g \bm g g都在点 x \bm x x处可微， u u u是在 x \bm x x处可微的数量值函数，则
(1) f + g \bm f+\bm g f+g在 x x x处可微，且其导数为 D ( f + g ) ( x ) = D f ( x ) + D g ( 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 x x处可微，且其导数为 D ⟨ f , g ⟩ ( x ) = ( f ( x ) ) T D g ( x ) + ( g ( x ) ) T D f ( 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) u f u\bm f uf在 x \bm x x处可微，且其导数为 D ( u f ) ( x ) = u ( x ) D f ( 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 → R 3 , g : R → R 3 \bm f:\mathbb{R}\to\mathbb{R}^3,\bm g:\mathbb{R}\to\mathbb{R}^3 f:R→R3,g:R→R3，则向量积 f × g \bm f\times\bm g f×g在 x x x处可微，且其导数为 D ( f × g ) ( x ) = D f ( x ) × g ( x ) + f ( x ) × D g ( 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 = [ f 1 f 2 ⋮ f m ] , g = [ g 1 g 2 ⋮ g m ] \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=
f1f2⋮fm
,g=
g1g2⋮gm
，则数量值函数 F = ⟨ f , g ⟩ ( x ) = ∑ i = 1 m f i g i F=\left\langle\bm f,\bm g\right\rangle(\bm x)=\sum\limits_{i=1}^mf_ig_i F=⟨f,g⟩(x)=i=1∑mfigi，且 D F ( x ) = ∇ F ( x ) = ∑ i = 1 m ∇ f i g i ( 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∇figi(x)我们知道，对于 j ∈ { 1 , 2 , ⋯ , n } j\in\{1,2,\cdots,n\} j∈{1,2,⋯,n}， ∂ f i g i ( x j ) ∂ x j = f i ∂ g i ( x j ) ∂ x j + g i ∂ f i ( x j ) ∂ x j \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∂figi(xj)=fi∂xj∂gi(xj)+gi∂xj∂fi(xj)故 ∇ f i g i ( x ) = ( f i ∂ g i ( x 1 ) ∂ x 1 + g i ∂ f i ( x 1 ) ∂ x 1 , f i ∂ g i ( x 2 ) ∂ x 2 + g i ∂ f i ( x 2 ) ∂ x 2 , ⋯ , f i ∂ g i ( x n ) ∂ x n + g i ∂ f i ( x n ) ∂ x n ) = f i ( x ) ∇ g i ( x ) + ∇ f i ( x ) g i ( 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} ∇figi(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 D F ( x ) = ∑ i = 1 m [ f i ( x ) ∇ g i ( x ) + ∇ f i ( x ) g i ( x ) ] = [ f 1 ( x ) f 2 ( x ) ⋯ f m ( x ) ] [ ∇ g 1 ( x ) ∇ g 2 ( x ) ⋮ ∇ g m ( x ) ] + [ g 1 ( x ) g 2 ( x ) ⋯ g m ( x ) ] [ ∇ f 1 ( x ) ∇ f 2 ( x ) ⋮ ∇ f m ( x ) ] = ( f ( x ) ) T D g ( x ) + ( g ( x ) ) T D f ( 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 ∣ i j k f 1 f 2 f 3 g 1 g 2 g 3 ∣ = d d x [ f 2 g 3 − f 3 g 2 f 3 g 1 − f 1 g 3 f 1 g 2 − f 2 g 1 ] = [ f 2 ′ g 3 + f 2 g 3 ′ − f 3 ′ g 2 − f 3 g 2 ′ f 3 ′ g 1 + f 3 g 1 ′ − f 1 ′ g 3 − f 1 g 3 ′ f 1 ′ g 2 + f 1 g 2 ′ − f 2 ′ g 1 − f 2 g 1 ′ ] = [ f 2 ′ g 3 − f 3 ′ g 2 f 3 ′ g 1 − f 1 ′ g 3 f 1 ′ g 2 − f 2 ′ g 1 ] + [ f 2 g 3 ′ − f 3 g 2 ′ f 3 g 1 ′ − f 1 g 3 ′ f 1 g 2 ′ − f 2 g 1 ′ ] = [ f 1 ′ f 2 ′ f 3 ′ ] × [ g 1 g 2 g 3 ] + [ f 1 f 2 f 3 ] × [ g 1 ′ g 2 ′ g 3 ′ ] = D f ( x ) × g ( x ) + f ( x ) × D g ( 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
if1g1jf2g2kf3g3
=dxd
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)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 c c c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 d d t ⟨ 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 d d t ∥ 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\rangle dtd∥r(t)∥2=2⟨r′(t),r(t)⟩故 ⟨ r ′ ( t ) , r ( t ) ⟩ ≡ 0 ⟺ d d t ∥ 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 ) : R n → R p \bm g(\bm x):\mathbb{R}^n\to\mathbb{R}^p g(x):Rn→Rp在点 x 0 \bm x_0 x0处可微，向量值函数 f ( u ) : R p → R m \bm f(\bm u):\mathbb{R}^p\to\mathbb{R}^m f(u):Rp→Rm在点 g ( x 0 ) \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 x 0 \bm x_0 x0处可微，且 D w ( x 0 ) = D f ( g ( x 0 ) ) D g ( x 0 ) \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， D w ( x ) \rm D\bm w(\bm x) Dw(x)的第 ( i , j ) (i,j) (i,j)个元素 ∂ w i ∂ x j \frac{\partial w_i}{\partial x_j} ∂xj∂wi可以表示为 ∂ w i ∂ x j = ∂ f i ( g ( x ) ) ∂ x j = ∂ f i ( g 1 ( x 1 , x 2 , ⋯ , x n ) , g 2 ( x 1 , x 2 , ⋯ , x n ) , ⋯ , g p ( x 1 , x 2 , ⋯ , x n ) ) ∂ x j = ∑ k = 1 p ∂ f i ∂ u k ∂ g k ∂ x j \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 = m n=p=m n=p=m时，取两端的行列式得 ∂ ( w 1 , w 2 , ⋯ , w n ) ∂ ( x 1 , x 2 , ⋯ , x n ) = ∂ ( w 1 , w 2 , ⋯ , w n ) ∂ ( u 1 , u 2 , ⋯ , u n ) ∂ ( g 1 , g 2 , ⋯ , g n ) ∂ ( x 1 , x 2 , ⋯ , x n ) \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（ J w = J f J g \bm J_w=\bm J_f\bm J_g Jw=JfJg）。
推论 设 f , g : R n → R n \bm f,\bm g:\mathbb{R}^n\to\mathbb{R}^n f,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 x w(x)=f(g(x))≡x，而 D w ( x ) = I \rm D\bm w(\bm x)=\bm I Dw(x)=I，故 J w = 1 \bm J_w=1 Jw=1， J f J g = 1 \bm J_f\bm J_g=1 JfJg=1。∎
三、由方程所确定的隐函数的微分法
例1 已知方程组 { F 1 ( x , y , u , v ) = x u + y v = 0 F 2 ( x , y , u , v ) = y u + x v = 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。
解：
(1) 方程组两边对 x x x求偏导得 { u + x ∂ u ∂ x + y ∂ v ∂ x = 0 y ∂ 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 [ x y y x ] [ ∂ 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} [xyyx][∂x∂u∂x∂v]=[−u−v]其中系数行列式即为雅可比式 J = ∂ ( F 1 , F 2 ) ∂ ( 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 = J u J = ∣ − u y − v x ∣ ∣ x y y x ∣ = v y − u x x 2 − y 2 \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=
xyyx
−u−vyx
=x2−y2vy−ux(2) 同理可得 ∂ v ∂ y = v y − u x x 2 − y 2 \frac{\partial v}{\partial y}=\frac{vy-ux}{x^2-y^2} ∂y∂v=x2−y2vy−ux。注意，原方程中将 u u u与 v v v互换、 x x x与 y y y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 = x u v + v w + w u = y u v w = 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。
解：两边取微分得 { d u + d v + d w = d x ( v + w ) d u + ( u + w ) d v + ( u + v ) d w = d y v w d u + u w d v + u v d w = d z \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 [ 1 1 1 v + w u + w u + v v w u w u v ] [ d u d v d w ] = [ d x d y d z ] \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+wvw1u+wuw1u+vuv
dudvdw
=
dxdydz
雅可比行列式 J = ∣ 1 1 1 v + w u + w u + v v w u w u v ∣ = u 2 ( v − w ) + v 2 ( w − u ) + w 2 ( 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+wvw1u+wuw1u+vuv
=u2(v−w)+v2(w−u)+w2(u−v)=(u−v)(v−w)(u−w)故 d u = J u J = ∣ d x 1 1 d y u + w u + v d z u w u v ∣ J = u 2 ( v − w ) d x − u ( v − w ) d y + ( v − w ) d z J = u 2 d x − u d y + d z ( 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
dxdydz1u+wuw1u+vuv
=Ju2(v−w)dx−u(v−w)dy+(v−w)dz=(u−v)(u−w)u2dx−udy+dz因此 { ∂ u ∂ x = u 2 ( 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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