矢量点积与矢量叉乘的微分
在对矢量点积与叉乘的微分公式进行推导之前，我们先看看这两个公式长什么样
矢量点积的微分：
d ( F ⃗ ⋅ x ⃗ ) = x ⃗ d F ⃗ + F ⃗ d x ⃗ (1) d(\vec{F}\cdot \vec{x}) = \vec{x}d\vec{F} + \vec{F}d\vec{x} \tag{1} d(F⋅x)=xdF+Fdx(1)
矢量叉乘的微分：
d ( r ⃗ × p ⃗ ) = d r ⃗ × p ⃗ + r ⃗ × d p ⃗ (2) d(\vec{r} \times \vec{p}) = d \vec{r} \times \vec{p} + \vec{r} \times d\vec{p} \tag{2} d(r×p)=dr×p+r×dp(2)
我们可以发现，其形式与多元函数的全微分形式一致，那么问题来了，为什么点积与叉乘也满足多元函数的链导法则？
首先，我们将矢量写成坐标形式来进行讨论.
对于矢量的点积，设 F ⃗ = ( x 1 , y 1 ) , x ⃗ = ( x 2 , y 2 ) \vec{F} = (x_1,y_1), \vec{x} = (x2,y2) F=(x1,y1),x=(x2,y2)，则有
d ( F ⃗ ⋅ x ⃗ ) = d ( x 1 x 2 + y 1 y 2 ) = d ( x 1 x 2 ) + d ( y 1 y 2 ) = x 2 d x 1 + x 1 d x 2 + y 2 d y 1 + y 1 d y 2 = ( x 2 , y 2 ) ( d x 1 , d y 1 ) + ( x 1 , y 1 ) ( d x 2 , d y 2 ) = x ⃗ d F ⃗ + F ⃗ d x ⃗ \begin{aligned}d(\vec{F}\cdot \vec{x}) &= d(x_1x_2 +y_1y_2) = d(x_1x_2) + d(y_1y_2) \\ \\&=x_2dx_1 + x_1dx_2 + y_2dy_1 + y_1dy_2 \\ \\&= (x_2, y_2)(dx_1, dy_1) + (x_1, y_1)(dx_2, dy_2) \\ \\&= \vec{x}d\vec{F} + \vec{F}d\vec{x}\end{aligned} d(F⋅x)=d(x1x2+y1y2)=d(x1x2)+d(y1y2)=x2dx1+x1dx2+y2dy1+y1dy2=(x2,y2)(dx1,dy1)+(x1,y1)(dx2,dy2)=xdF+Fdx
同样，对于矢量的叉乘，设 r ⃗ = ( x 1 , y 1 , z 1 ) , p ⃗ = ( x 2 , y 2 , z 2 ) \vec{r} = (x_1, y_1, z_1), \vec{p} = (x_2, y_2, z_2) r=(x1,y1,z1),p=(x2,y2,z2)，则有
r ⃗ × p ⃗ = ∣ i ⃗ j ⃗ k ⃗ x 1 y 1 z 1 x 2 y 2 z 2 ∣ = ∣ y 1 z 1 y 2 z 2 ∣ i ⃗ − ∣ x 1 z 1 x 2 z 2 ∣ j ⃗ + ∣ x 1 y 1 x 2 y 2 ∣ k ⃗ \begin{aligned} \vec{r} \times \vec{p} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \end{vmatrix} = \begin{vmatrix} y_1 & z_1 \\ y_2 & z_2 \end{vmatrix} \ \vec{i} - \begin{vmatrix} x_1 & z_1 \\ x_2 & z_2 \end{vmatrix} \ \vec{j} + \begin{vmatrix} x_1 & y_1 \\ x_2 & y_2 \end{vmatrix} \ \vec{k}\end{aligned} r×p=∣∣∣∣∣∣ix1x2jy1y2kz1z2∣∣∣∣∣∣=∣∣∣∣y1y2z1z2∣∣∣∣ i−∣∣∣∣x1x2z1z2∣∣∣∣ j+∣∣∣∣x1x2y1y2∣∣∣∣ k
所以
d ( r ⃗ × p ⃗ ) = d ∣ y 1 z 1 y 2 z 2 ∣ i ⃗ − d ∣ x 1 z 1 x 2 z 2 ∣ j ⃗ + d ∣ x 1 y 1 x 2 y 2 ∣ k ⃗ = d ( y 1 z 2 − z 1 y 2 ) i ⃗ − d ( x 1 z 2 − z 1 x 2 ) j ⃗ + d ( x 1 y 2 − y 1 x 2 ) k ⃗ = ( z 2 d y 1 + y 1 d z 2 − y 2 d z 1 − z 1 d y 2 ) i ⃗ − ( z 2 d x 1 + x 1 d z 2 − x 2 d z 1 − z 1 d x 2 ) j ⃗ + ( y 2 d y 1 + x 1 d y 2 − x 2 d y 1 − y 1 d x 2 ) k ⃗ \begin{aligned}d(\vec{r} \times \vec{p}) &= d\begin{vmatrix} y_1 & z_1 \\ y_2 & z_2 \end{vmatrix} \ \vec{i} - d\begin{vmatrix} x_1 & z_1 \\ x_2 & z_2 \end{vmatrix} \ \vec{j} + d\begin{vmatrix} x_1 & y_1 \\ x_2 & y_2 \end{vmatrix} \ \vec{k} \\ \\&=d(y_1z_2 - z_1y_2)\ \vec{i} -d(x_1z_2 - z_1x_2) \ \vec{j} + d(x_1y_2 - y_1x_2) \ \vec{k} \\ \\ &= (z_2dy_1 + y_1dz_2 - y_2dz_1 - z_1dy_2)\ \vec{i} - (z_2dx_1 + x_1dz_2 - x_2dz_1 - z_1dx_2) \ \vec{j}\\\\ &+ (y_2dy_1 + x_1dy_2 - x_2dy_1 - y_1dx_2) \ \vec{k}\end{aligned} d(r×p)=d∣∣∣∣y1y2z1z2∣∣∣∣ i−d∣∣∣∣x1x2z1z2∣∣∣∣ j+d∣∣∣∣x1x2y1y2∣∣∣∣ k=d(y1z2−z1y2) i−d(x1z2−z1x2) j+d(x1y2−y1x2) k=(z2dy1+y1dz2−y2dz1−z1dy2) i−(z2dx1+x1dz2−x2dz1−z1dx2) j+(y2dy1+x1dy2−x2dy1−y1dx2) k
又因为
d r ⃗ = ( d x 1 , d y 1 , d z 1 ) ， d p ⃗ = ( d x 2 , d y 2 , d z 2 ) \begin{aligned}d\vec{r} &= (dx_1, dy_1, dz_1)，d\vec{p} = (dx_2, dy_2, dz_2)\end{aligned} dr=(dx1,dy1,dz1)，dp=(dx2,dy2,dz2)
故
d r ⃗ × p ⃗ + r ⃗ × d p ⃗ = ∣ i ⃗ j ⃗ k ⃗ d x 2 d y 2 d z 2 x 1 y 1 z 1 ∣ + ∣ i ⃗ j ⃗ k ⃗ x 2 y 2 z 2 d x 1 d y 1 d z 1 ∣ = ( z 2 d y 1 + y 1 d z 2 − y 2 d z 1 − z 1 d y 2 ) i ⃗ − ( z 2 d x 1 + x 1 d z 2 − x 2 d z 1 − z 1 d x 2 ) j ⃗ + ( y 2 d y 1 + x 1 d y 2 − x 2 d y 1 − y 1 d x 2 ) k ⃗ = d ( r ⃗ × p ⃗ ) \begin{aligned}d\vec{r} \times\vec{p} + \vec{r} \times d\vec{p} &= \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ dx_2 & dy_2 & dz_2 \\ x_1 & y_1 & z_1 \end{vmatrix} + \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ x_2 & y_2 & z_2 \\ dx_1 & dy_1 & dz_1 \end{vmatrix} \\ \\ &= (z_2dy_1 + y_1dz_2 - y_2dz_1 - z_1dy_2)\ \vec{i} - (z_2dx_1 + x_1dz_2 - x_2dz_1 - z_1dx_2) \ \vec{j}\\\\ &+ (y_2dy_1 + x_1dy_2 - x_2dy_1 - y_1dx_2) \ \vec{k}\\ \\&= d(\vec{r} \times \vec{p})\end{aligned} dr×p+r×dp=∣∣∣∣∣∣idx2x1jdy2y1kdz2z1∣∣∣∣∣∣+∣∣∣∣∣∣ix2dx1jy2dy1kz2dz1∣∣∣∣∣∣=(z2dy1+y1dz2−y2dz1−z1dy2) i−(z2dx1+x1dz2−x2dz1−z1dx2) j+(y2dy1+x1dy2−x2dy1−y1dx2) k=d(r×p)
坐标角度进行推导的过程未免太过麻烦，那么我们不如换一个方向，直接从全微分的定义来进行推导
根据全微分的定义：
设函数 z = f ( x , y ) z = f(x, y) z=f(x,y) 在点 ( x 0 , y 0 ) (x_0, y_0) (x0,y0) 的某邻域内有定义， P ′ ( x + Δ x , y + Δ y ) P'(x + \Delta x, y + \Delta y) P′(x+Δx,y+Δy) 为该邻域内任意一点，则称
Δ z = f ( x + Δ x , y + Δ y ) − f ( x , y ) \Delta z = f(x + \Delta x, y + \Delta y) - f(x, y) Δz=f(x+Δx,y+Δy)−f(x,y)
为函数在点 P ( x , y ) P(x, y) P(x,y) 处的全增量.
若函数 z = f ( x , y ) z = f(x, y) z=f(x,y) 在点 P ( x , y ) P(x, y) P(x,y) 处的全增量能表示为
Δ z = A Δ x + B Δ y + o ( ρ ) \Delta z = A\Delta x + B\Delta y + o(\rho) Δz=AΔx+BΔy+o(ρ)
的形式，则说函数 z = f ( x , y ) z = f(x, y) z=f(x,y) 在点 P P P 处可微，并称 A Δ x + B Δ y A\Delta x + B \Delta y AΔx+BΔy 为函数在点 P P P 处的全微分
那么我们先来试试矢量的点积
lim Δ F ⃗ → ( 0 , 0 ) , Δ x ⃗ → ( 0 , 0 ) Δ ( F ⃗ ⋅ x ⃗ ) = lim Δ F ⃗ → ( 0 , 0 ) , Δ x ⃗ → ( 0 , 0 ) ( F ⃗ + Δ F ⃗ ) ⋅ ( x ⃗ + Δ x ⃗ ) − F ⃗ ⋅ x ⃗ = Δ F ⃗ ⋅ x ⃗ + F ⃗ ⋅ Δ x ⃗ + Δ F ⃗ ⋅ Δ x ⃗ = F ⃗ ⋅ Δ x ⃗ + x ⃗ ⋅ Δ F ⃗ + o ( ρ ) \begin{aligned} \lim_{\Delta \vec{F}\to (0, 0) , \Delta \vec{x} \to (0, 0)} \Delta(\vec{F}\cdot \vec{x}) &= \lim_{\Delta \vec{F} \to (0, 0), \Delta \vec{x} \to (0, 0)}(\vec{F} + \Delta \vec{F}) \cdot (\vec{x} + \Delta\vec{x}) - \vec{F} \cdot \vec{x} \\ \\ &= \Delta \vec{F} \cdot \vec{x} + \vec{F} \cdot \Delta \vec{x} + \Delta \vec{F} \cdot \Delta\vec{x} \\ \\ &= \vec{F} \cdot \Delta \vec{x} + \vec{x} \cdot \Delta \vec{F} + o(\rho) \end{aligned} ΔF→(0,0),Δx→(0,0)limΔ(F⋅x)=ΔF→(0,0),Δx→(0,0)lim(F+ΔF)⋅(x+Δx)−F⋅x=ΔF⋅x+F⋅Δx+ΔF⋅Δx=F⋅Δx+x⋅ΔF+o(ρ)
所以，我们可以得到
d ( F ⃗ ⋅ x ⃗ ) = x ⃗ d F ⃗ + F ⃗ d x ⃗ d(\vec{F}\cdot \vec{x}) = \vec{x}d\vec{F} + \vec{F}d\vec{x} d(F⋅x)=xdF+Fdx
接下来，我们尝试推导矢量的叉乘
lim Δ r ⃗ → ( 0 , 0 , 0 ) , Δ p ⃗ → ( 0 , 0 , 0 ) Δ ( r ⃗ × p ⃗ ) = lim Δ r ⃗ → ( 0 , 0 , 0 ) , Δ p ⃗ → ( 0 , 0 , 0 ) ( r ⃗ + Δ r ⃗ ) × ( p ⃗ + Δ p ⃗ ) − r ⃗ × p ⃗ = r ⃗ × Δ p ⃗ + Δ r ⃗ × p ⃗ + Δ r ⃗ × Δ p ⃗ = r ⃗ × Δ p ⃗ + Δ r ⃗ × p ⃗ + o ( ρ ) \begin{aligned}\lim\limits_{\Delta \vec{r} \to (0, 0, 0) , \Delta \vec p \to (0, 0, 0)}\Delta(\vec{r} \times \vec{p}) &= \lim\limits_{\Delta \vec{r} \to (0, 0, 0) , \Delta \vec p \to (0, 0, 0)} (\vec{r} + \vec{\Delta r}) \times (\vec{p} + \Delta \vec{p}) - \vec{r} \times \vec{p} \\ \\&= \vec{r} \times \vec{\Delta p} + \Delta \vec{r} \times \vec{p} + \vec{\Delta r} \times \vec{\Delta p} \\ \\&= \vec{r} \times \vec{\Delta p} + \Delta \vec{r} \times \vec{p} + o(\rho)\end{aligned} Δr→(0,0,0),Δp→(0,0,0)limΔ(r×p)=Δr→(0,0,0),Δp→(0,0,0)lim(r+Δr)×(p+Δp)−r×p=r×Δp+Δr×p+Δr×Δp=r×Δp+Δr×p+o(ρ)
所以，我们同样可以得到
d ( r ⃗ × p ⃗ ) = d r ⃗ × p ⃗ + r ⃗ × d p ⃗ d(\vec{r} \times \vec{p}) = d \vec{r} \times \vec{p} + \vec{r} \times d\vec{p} d(r×p)=dr×p+r×dp
与此同时，我们注意到（1）式与（2）式之间细微的区别，将两个等式放在一起比较
d ( F ⃗ ⋅ x ⃗ ) = x ⃗ d F ⃗ + F ⃗ d x ⃗ d ( r ⃗ × p ⃗ ) = d r ⃗ × p ⃗ + r ⃗ × d p ⃗ \begin{aligned}d(\vec{F}\cdot \vec{x}) &= \vec{x}d\vec{F} + \vec{F}d\vec{x} \\ \\d(\vec{r} \times \vec{p}) &= d \vec{r} \times \vec{p} + \vec{r} \times d\vec{p}\end{aligned} d(F⋅x)d(r×p)=xdF+Fdx=dr×p+r×dp
不难发现，由于矢量的点积满足交换律，而叉乘不满足交换律，矢量点积的等号右边全部可以写成 x 1 d y 1 + y 1 d x 1 x_1dy_1 + y_1dx_1 x1dy1+y1dx1 的形式，而矢量叉乘必须保持原有顺序， d x dx dx 部分与 y y y 部分不能交换顺序。
