矩阵微分笔记（2）

前言

这篇笔记的内容是基于参考的文章写出的，公式部分可以会沿用文章本来的式，但会加入我自己的一些思考以及注释，如果读者认为我写的不够好得话可以参考原文章~

本笔记的内容是学习向量变元的实值标量函数、矩阵变元的实值标量函数中最基础的矩阵求导公式(会对个别重要的公式做证明)。

下面有一个求矩阵导数的网站，可以用来验证求导结果是否正确：Matrix Calculus

基本求导规则

1. 向量变元的实值标量函数

即形如$$f(x),x=[x_1,x_2,\cdots ,x_n{]}^T$$使用梯度向量形式，则有$${\nabla }_{x}f(x)=\frac{\partial f(x)}{\partial x}={\left[\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2},\cdots ,\frac{\partial f}{\partial x_n}\right]}^T$$对于该形式的求导法则，与高等数学中的导数的法则的证明思想类似，下面给个4个原则，并选择性给出证明：

1.1 4个法则

（1）：常数求导

与一元函数常数求导相同：结果为零向量，即

$$\frac{\partial c}{\partial x}=0_{n\times 1}$$其中，$c$ 为常数

（2）：线性法则

与一元函数求导线性法则相同：相加再求导等于求导再相加，常数提到外面，即：$$\frac{\partial [c_{1}f(x)+c_{2}g(x)]}{\partial x}=c_1\frac{\partial f(x)}{\partial x}+c_2\frac{\partial g(x)}{\partial x}$$其中，$c_1,c_2$ 为常数。

（3）：乘积法则

与一元函数求导乘积法则相同：前导后不导加前不导后导，即

$$\frac{\partial [f(x)g(x)]}{\partial x}=\frac{\partial f(x)}{\partial x}g(x)+f(x)\frac{\partial g(x)}{\partial x}$$证明： ∂ [ f ( x ⃗ ) g ( x ⃗ ) ] ∂ x ⃗ = [ ∂ ( f g ) ∂ x 1 ∂ ( f g ) ∂ x 2 ⋮ ∂ ( f g ) ∂ x n ] = [ ∂ f ∂ x 1 g + f ∂ g ∂ x 1 ∂ f ∂ x 2 g + f ∂ g ∂ x 2 ⋮ ∂ f ∂ x n g + f ∂ g ∂ x n ] = [ ∂ f ∂ x 1 ∂ f ∂ x 2 ⋮ ∂ f ∂ x n ] g + f [ ∂ g ∂ x 1 ∂ g ∂ x 2 ⋮ ∂ g ∂ x n ] = ∂ f ( x ⃗ ) ∂ x ⃗ g ( x ⃗ ) + f ( x ⃗ ) ∂ g ( x ⃗ ) ∂ x ⃗ \begin{aligned} \frac{\partial[f(\vec{x})g(\vec{x})]}{\partial \vec{x}}& =\begin{bmatrix}\frac{\partial(fg)}{\partial x_1}\\\frac{\partial(fg)}{\partial x_2}\\\vdots\\\frac{\partial(fg)}{\partial x_n}\end{bmatrix} \\ &=\begin{bmatrix}\frac{\partial f}{\partial x_1}g+f\frac{\partial g}{\partial x_1}\\\frac{\partial f}{\partial x_2}g+f\frac{\partial g}{\partial x_2}\\\vdots\\\frac{\partial f}{\partial x_n}g+f\frac{\partial g}{\partial x_n}\end{bmatrix} \\ &\left.=\left[\begin{array}{c}\frac{\partial f}{\partial x_1}\\\frac{\partial f}{\partial x_2}\\\vdots\\\frac{\partial f}{\partial x_n}\end{array}\right.\right]g+f\left[\begin{array}{c}\frac{\partial g}{\partial x_1}\\\frac{\partial g}{\partial x_2}\\\vdots\\\frac{\partial g}{\partial x_n}\end{array}\right] \\ &=\frac{\partial f(\vec{x})}{\partial\vec{x}}g(\vec{x})+f(\vec{x})\frac{\partial g(\vec{x})}{\partial\vec{x}} \end{aligned} ∂x ∂[f(x )g(x )]​​= ​∂x1​∂(fg)​∂x2​∂(fg)​⋮∂xn​∂(fg)​​ ​= ​∂x1​∂f​g+f∂x1​∂g​∂x2​∂f​g+f∂x2​∂g​⋮∂xn​∂f​g+f∂xn​∂g​​ ​= ​∂x1​∂f​∂x2​∂f​⋮∂xn​∂f​​ ​g+f ​∂x1​∂g​∂x2​∂g​⋮∂xn​∂g​​ ​=∂x ∂f(x )​g(x )+f(x )∂x ∂g(x )​​

（4）：商法则

与一元函数求导商法则相同：（上导下不导 减 上不导下导）除以（下的平方）：$$\begin{array}{rl}& \frac{\partial \left[\frac{f(x)}{g(x)}\right]}{\partial x}=\frac{1}{g^2(x)}\left[\frac{\partial f(x)}{\partial x}g(x)-f(x)\frac{\partial g(x)}{\partial x}\right]\end{array}$$其中，$g(x)\ne 0$

证明： ∂ [ f ( x ⃗ ) g ( x ⃗ ) ] ∂ x ⃗ = [ ∂ ( f g ) ∂ x 1 ∂ ( f g ) ∂ x 2 ⋮ ∂ ( f g ) ∂ x n ] = [ 1 g 2 ( ∂ f ∂ x 1 g − f ∂ g ∂ x 1 ) 1 g 2 ( ∂ f ∂ x 2 g − f ∂ g ∂ x 2 ) ⋮ 1 g 2 ( ∂ f ∂ x n g − f ∂ g ∂ x n ) ] = 1 g 2 ( [ ∂ f ∂ x 1 ∂ f ∂ x 2 ⋮ ∂ f ∂ x n ] g − f [ ∂ g ∂ x 1 ∂ g ∂ x 2 ⋮ ∂ g ∂ x n ] ) = 1 g 2 ( x ⃗ ) [ ∂ f ( x ⃗ ) ∂ x ⃗ g ( x ⃗ ) − f ( x ⃗ ) ∂ g ( x ⃗ ) ∂ x ⃗ ] \begin{aligned} \frac{\partial\left[\frac{f(\vec{x})}{g(\vec{x})}\right]}{\partial\vec{x}}& \left.=\left[\begin{array}{c}\dfrac{\partial(\frac{f}{g})}{\partial x_1}\\\dfrac{\partial(\frac{f}{g})}{\partial x_2}\\\vdots\\\dfrac{\partial(\frac{f}{g})}{\partial x_n}\end{array}\right.\right] \\ &=\begin{bmatrix}\frac{1}{g^2}\left(\frac{\partial f}{\partial x_1}g-f\frac{\partial g}{\partial x_1}\right)\\\frac{1}{g^2}\left(\frac{\partial f}{\partial x_2}g-f\frac{\partial g}{\partial x_2}\right)\\\vdots\\\frac{1}{g^2}\left(\frac{\partial f}{\partial x_n}g-f\frac{\partial g}{\partial x_n}\right)\end{bmatrix} \\ &\left.\left.=\frac{1}{g^2}\left(\left[\begin{array}{c}\frac{\partial f}{\partial x_1}\\\frac{\partial f}{\partial x_2}\\\vdots\\\frac{\partial f}{\partial x_n}\end{array}\right.\right.\right]g-f\left[\begin{array}{c}\frac{\partial g}{\partial x_1}\\\frac{\partial g}{\partial x_2}\\\vdots\\\frac{\partial g}{\partial x_n}\end{array}\right]\right) \\ &=\frac{1}{g^{2}(\vec{x})}\left[\frac{\partial f(\vec{x})}{\partial\vec{x}}g(\vec{x})-f(\vec{x})\frac{\partial g(\vec{x})}{\partial\vec{x}}\right] \end{aligned} ∂x ∂[g(x )f(x )​]​​= ​∂x1​∂(gf​)​∂x2​∂(gf​)​⋮∂xn​∂(gf​)​​ ​= ​g21​(∂x1​∂f​g−f∂x1​∂g​)g21​(∂x2​∂f​g−f∂x2​∂g​)⋮g21​(∂xn​∂f​g−f∂xn​∂g​)​ ​=g21​ ​ ​∂x1​∂f​∂x2​∂f​⋮∂xn​∂f​​ ​g−f ​∂x1​∂g​∂x2​∂g​⋮∂xn​∂g​​ ​ ​=g2(x )1​[∂x ∂f(x )​g(x )−f(x )∂x ∂g(x )​]​如上所述，证明完毕

1.2 常用公式

（1）：
$$\frac{\partial (x^{T}a)}{\partial x}=\frac{\partial (a^{T}x)}{\partial x}=a$$

其中，$a$ 为常数向量，即 $a=(a_1,a_2,\cdots ,a_n{)}^T$。

证明：该式采用是是向量变元对标量函数的分布布局，结果如下： ∂ ( x ⃗ T a ⃗ ) ∂ x = ∂ ( a ⃗ T x ⃗ ) ∂ x ⃗ = ∂ ( a 1 x 1 + a 2 x 2 + ⋯ + a n x n ) ∂ x ⃗ = [ ∂ ( a 1 x 1 + a 2 x 2 + ⋯ + a n x n ) ∂ x 1 ∂ ( a 1 x 1 + a 2 x 2 + ⋯ + a n x n ) ∂ x 2 ⋮ ∂ ( a 1 x 1 + a 2 x 2 + ⋯ + a n x n ) ∂ x n ] = [ a 1 a 2 ⋮ a n ] = a ⃗ \begin{aligned} \frac{\partial(\vec{x}^{T}\vec{a})}{\partial x}& =\frac{\partial(\vec{a}^T\vec{x})}{\partial\vec{x}} \\ &=\frac{\partial(a_1x_1+a_2x_2+\cdots+a_nx_n)}{\partial\vec{x}} \\ &\left.=\left[\begin{array}{c}\frac{\partial(a_1x_1+a_2x_2+\cdots+a_nx_n)}{\partial x_1}\\\frac{\partial(a_1x_1+a_2x_2+\cdots+a_nx_n)}{\partial x_2}\\\vdots\\\frac{\partial(a_1x_1+a_2x_2+\cdots+a_nx_n)}{\partial x_n}\end{array}\right.\right] \\ &\left.=\left[\begin{array}{c}a_1\\a_2\\\vdots\\a_n\end{array}\right.\right] \\ &=\vec{a} \end{aligned} ∂x∂(x Ta )​​=∂x ∂(a Tx )​=∂x ∂(a1​x1​+a2​x2​+⋯+an​xn​)​= ​∂x1​∂(a1​x1​+a2​x2​+⋯+an​xn​)​∂x2​∂(a1​x1​+a2​x2​+⋯+an​xn​)​⋮∂xn​∂(a1​x1​+a2​x2​+⋯+an​xn​)​​ ​= ​a1​a2​⋮an​​ ​=a ​

（2）：$$\frac{\partial (x^{T}x)}{\partial x}=2x$$证明：该式采用是是向量变元对标量函数的分布布局，结果如下： ∂ ( x ⃗ T x ⃗ ) ∂ x ⃗ = ∂ ( x 1 2 + x 2 2 + ⋯ + x n 2 ) ∂ x ⃗ = [ ∂ ( x 1 2 + x 2 2 + ⋯ + x n 2 ) ∂ x 1 ∂ ( x 1 2 + x 2 2 + ⋯ + x n 2 ) ∂ x 2 ⋮ ∂ ( x 1 2 + x 2 2 + ⋯ + x n 2 ) ∂ x n ] = [ 2 x 1 2 x 2 ⋮ 2 x n ] = 2 [ x 1 x 2 ⋮ x n ] = 2 x ⃗ \begin{aligned} \frac{\partial(\vec{x}^{T}\vec{x})}{\partial \vec{x}}& =\frac{\partial(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2})}{\partial\vec{x}} \\ &\left.=\left[\begin{array}{c}\frac{\partial(x_1^2+x_2^2+\cdots+x_n^2)}{\partial x_1}\\\frac{\partial(x_1^2+x_2^2+\cdots+x_n^2)}{\partial x_2}\\\vdots\\\frac{\partial(x_1^2+x_2^2+\cdots+x_n^2)}{\partial x_n}\end{array}\right.\right] \\ &=\begin{bmatrix}2x_1\\2x_2\\\vdots\\2x_n\end{bmatrix} \\ &\left.=2\left[\begin{array}{c}x_1\\x_2\\\vdots\\x_n\end{array}\right.\right] \\ &=2\vec{x} \end{aligned} ∂x ∂(x Tx )​​=∂x ∂(x12​+x22​+⋯+xn2​)​= ​∂x1​∂(x12​+x22​+⋯+xn2​)​∂x2​∂(x12​+x22​+⋯+xn2​)​⋮∂xn​∂(x12​+x22​+⋯+xn2​)​​ ​= ​2x1​2x2​⋮2xn​​ ​=2 ​x1​x2​⋮xn​​ ​=2x ​

（3）：
$$\frac{\partial (x^{T}Ax)}{\partial x}=Ax+A^{T}x$$其中，$A_{n\times n}$是常数矩阵，$A_{n\times n}=(a_{ij}{)}_{i=1,j=1}^{n,n}$

证明： ∂ ( x ⃗ T A x ⃗ ) ∂ x ⃗ = ( x 1 , x 2 , … , x n ) ( a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋮ ⋮ ⋱ ⋮ a n 1 a n 2 ⋯ a n n ) ( x 1 x 2 ⋮ x n ) = ∂ ( a 11 x 1 x 1 + a 12 x 1 x 2 + ⋯ + a 1 n x 1 x n + a 21 x 2 x 1 + a 22 x 2 x 2 + ⋯ + a 2 n x 2 x n + ⋯ + a n 1 x n x 1 + a n 2 x n x 2 + ⋯ + a n n x n x n ) ∂ x ⃗ = [ ∂ ( a 11 x 1 x 1 + a 12 x 1 x 2 + ⋯ + a 1 n x 1 x n + a 21 x 2 x 1 + a 22 x 2 x 2 + ⋯ + a 2 n x 2 x n + ⋯ + a n 1 x n x 1 + a n 2 x n x 2 + ⋯ + a n n x n x n ) ∂ x 1 ∂ ( a 11 x 1 x 1 + a 12 x 1 x 2 + ⋯ + a 1 n x 1 x n + a 21 x 2 x 1 + a 22 x 2 x 2 + ⋯ + a 2 n x 2 x n + ⋯ + a n 1 x n x 1 + a n 2 x n x 2 + ⋯ + a n n x n x n ) ∂ x 2 ⋮ ∂ ( a 11 x 1 x 1 + a 12 x 1 x 2 + ⋯ + a 1 n x 1 x n + a 21 x 2 x 1 + a 22 x 2 x 2 + ⋯ + a 2 n x 2 x n + ⋯ + a n 1 x n x 1 + a n 2 x n x 2 + ⋯ + a n n x n x n ) ∂ x n ] = [ ( a 11 x 1 + a 12 x 2 + ⋯ + a 1 n x n ) + ( a 11 x 1 + a 21 x 2 + ⋯ + a n 1 x n ) ( a 21 x 1 + a 22 x 2 + ⋯ + a 2 n x n ) + ( a 12 x 1 + a 22 x 2 + ⋯ + a n 2 x n ) ⋮ ( a n 1 x 1 + a n 2 x 2 + ⋯ + a n n x n ) + ( a 1 n x 1 + a 2 n x 2 + ⋯ + a n n x n ) ] = [ a 11 x 1 + a 12 x 2 + ⋯ + a 1 n x n a 21 x 1 + a 22 x 2 + ⋯ + a 2 n x n ⋮ a n 1 x 1 + a n 2 x 2 + ⋯ + a n n x n ] + [ a 11 x 1 + a 21 x 2 + ⋯ + a n 1 x n a 12 x 1 + a 22 x 2 + ⋯ + a n 2 x n ⋮ a 1 n x 1 + a 2 n x 2 + ⋯ + a n n x n ] = [ a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋮ ⋮ ⋱ ⋮ a n 1 a n 2 ⋯ a n n ] [ x 1 x 2 ⋮ x n ] + [ a 11 a 21 ⋯ a n 1 a 12 a 22 ⋯ a n 2 ⋮ ⋮ ⋱ ⋮ a 1 n a 2 n ⋯ a n n ] [ x 1 x 2 ⋮ x n ] = A x ⃗ + A T x ⃗ (14) \begin{aligned} \frac{\partial( \vec{x}^T \pmb{A}\vec{x})}{\partial{\vec{x}}} &= (x_1,x_2,\dots,x_n)\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix} \\ &=\frac{\partial(a_{11}x_1x_1+a_{12}x_1x_2+\cdots+a_{1n}x_1x_n \\ +a_{21}x_2x_1+a_{22}x_2x_2+\cdots+a_{2n}x_2x_n \\ + \cdots \\ +a_{n1}x_nx_1+a_{n2}x_nx_2+\cdots+a_{nn}x_nx_n)}{\partial{\vec{x}}} \\\\ &= \begin{bmatrix} \frac{\partial(a_{11}x_1x_1+a_{12}x_1x_2+\cdots+a_{1n}x_1x_n \\ +a_{21}x_2x_1+a_{22}x_2x_2+\cdots+a_{2n}x_2x_n \\ + \cdots \\ +a_{n1}x_nx_1+a_{n2}x_nx_2+\cdots+a_{nn}x_nx_n)}{\partial{x_1}} \\ \frac{\partial(a_{11}x_1x_1+a_{12}x_1x_2+\cdots+a_{1n}x_1x_n \\ +a_{21}x_2x_1+a_{22}x_2x_2+\cdots+a_{2n}x_2x_n \\ + \cdots \\ +a_{n1}x_nx_1+a_{n2}x_nx_2+\cdots+a_{nn}x_nx_n)}{\partial{x_2}} \\ \vdots \\ \frac{\partial(a_{11}x_1x_1+a_{12}x_1x_2+\cdots+a_{1n}x_1x_n \\ +a_{21}x_2x_1+a_{22}x_2x_2+\cdots+a_{2n}x_2x_n \\ + \cdots \\ +a_{n1}x_nx_1+a_{n2}x_nx_2+\cdots+a_{nn}x_nx_n)}{\partial{x_n}} \end{bmatrix} \\\\ &= \begin{bmatrix} (a_{11}x_1+a_{12}x_2+\cdots+a_{1n}x_n)+(a_{11}x_1+a_{21}x_2+\cdots+a_{n1}x_n) \\ (a_{21}x_1+a_{22}x_2+\cdots+a_{2n}x_n)+(a_{12}x_1+a_{22}x_2+\cdots+a_{n2}x_n) \\ \vdots \\ (a_{n1}x_1+a_{n2}x_2+\cdots+a_{nn}x_n)+(a_{1n}x_1+a_{2n}x_2+\cdots+a_{nn}x_n) \end{bmatrix} \\\\ &= \begin{bmatrix} a_{11}x_1+a_{12}x_2+\cdots+a_{1n}x_n \\ a_{21}x_1+a_{22}x_2+\cdots+a_{2n}x_n \\ \vdots \\ a_{n1}x_1+a_{n2}x_2+\cdots+a_{nn}x_n \end{bmatrix} +\begin{bmatrix} a_{11}x_1+a_{21}x_2+\cdots+a_{n1}x_n \\ a_{12}x_1+a_{22}x_2+\cdots+a_{n2}x_n \\ \vdots \\ a_{1n}x_1+a_{2n}x_2+\cdots+a_{nn}x_n \end{bmatrix} \\\\ &= \begin{bmatrix} a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn} \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ \vdots\\ x_n \end{bmatrix} +\begin{bmatrix} a_{11}&a_{21}&\cdots&a_{n1}\\ a_{12}&a_{22}&\cdots&a_{n2}\\ \vdots&\vdots&\ddots&\vdots\\ a_{1n}&a_{2n}&\cdots&a_{nn} \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ \vdots\\ x_n \end{bmatrix} \\\\ &= \pmb{A}\vec{x}+\pmb{A}^T \vec{x} \end{aligned} \\\\ \tag{14} ∂x ∂(x TAx )​​=(x1​,x2​,…,xn​) ​a11​a21​⋮an1​​a12​a22​⋮an2​​⋯⋯⋱⋯​a1n​a2n​⋮ann​​ ​ ​x1​x2​⋮xn​​ ​=∂x ∂(a11​x1​x1​+a12​x1​x2​+⋯+a1n​x1​xn​+a21​x2​x1​+a22​x2​x2​+⋯+a2n​x2​xn​+⋯+an1​xn​x1​+an2​xn​x2​+⋯+ann​xn​xn​)​= ​∂x1​∂(a11​x1​x1​+a12​x1​x2​+⋯+a1n​x1​xn​+a21​x2​x1​+a22​x2​x2​+⋯+a2n​x2​xn​+⋯+an1​xn​x1​+an2​xn​x2​+⋯+ann​xn​xn​)​∂x2​∂(a11​x1​x1​+a12​x1​x2​+⋯+a1n​x1​xn​+a21​x2​x1​+a22​x2​x2​+⋯+a2n​x2​xn​+⋯+an1​xn​x1​+an2​xn​x2​+⋯+ann​xn​xn​)​⋮∂xn​∂(a11​x1​x1​+a12​x1​x2​+⋯+a1n​x1​xn​+a21​x2​x1​+a22​x2​x2​+⋯+a2n​x2​xn​+⋯+an1​xn​x1​+an2​xn​x2​+⋯+ann​xn​xn​)​​ ​= ​(a11​x1​+a12​x2​+⋯+a1n​xn​)+(a11​x1​+a21​x2​+⋯+an1​xn​)(a21​x1​+a22​x2​+⋯+a2n​xn​)+(a12​x1​+a22​x2​+⋯+an2​xn​)⋮(an1​x1​+an2​x2​+⋯+ann​xn​)+(a1n​x1​+a2n​x2​+⋯+ann​xn​)​ ​= ​a11​x1​+a12​x2​+⋯+a1n​xn​a21​x1​+a22​x2​+⋯+a2n​xn​⋮an1​x1​+an2​x2​+⋯+ann​xn​​ ​+ ​a11​x1​+a21​x2​+⋯+an1​xn​a12​x1​+a22​x2​+⋯+an2​xn​⋮a1n​x1​+a2n​x2​+⋯+ann​xn​​ ​= ​a11​a21​⋮an1​​a12​a22​⋮an2​​⋯⋯⋱⋯​a1n​a2n​⋮ann​​ ​ ​x1​x2​⋮xn​​ ​+ ​a11​a12​⋮a1n​​a21​a22​⋮a2n​​⋯⋯⋱⋯​an1​an2​⋮ann​​ ​ ​x1​x2​⋮xn​​ ​=Ax +ATx ​(14)上述的第一个等号是直接按照定义展开得到的，由于分子 $x^{T}Ax$ 实际上是一个标量，于是我们可以应用向量对标量函数的导数的求导法则来计算。通过观察我们可以发现结果布局是 $n\times 1$ 维的，且第一个分量的结果可以分成两项：是$A$ 的第一列的转置与 $x$ 的内积以及 $A$ 的第一行与 $x$ 的内积，为此可以很自然地计算剩余的分量

（4）：$$\frac{\partial (a^{T}x{x}^{T}b)}{\partial x}=a{b}^{T}x+b{a}^{T}x$$其中$a,b$ 为常数向量，$a=(a_1,a_2,\cdots ,a_n{)}^T,b=(b_1,b_2,\cdots ,b_n{)}^T$

证明：因为 $a^{T}x=x^{T}a,x^{T}b=b^{T}x$ ，所以有$$\frac{\partial (a^{T}x{x}^{T}b)}{\partial x}=\frac{\partial (x^{T}a{b}^{T}x)}{\partial x}$$因为 $a{b}^T$ 是常数矩阵，于是可以利用公式 $$\frac{\partial (x^{T}Ax)}{\partial x}=Ax+A^{T}x$$得到$$\frac{\partial (a^{T}x{x}^{T}b)}{\partial x}=\frac{\partial (x^{T}a{b}^{T}x)}{\partial x}=a{b}^{T}x+b{a}^{T}x$$如上所述，证明完毕

2. 矩阵变元的实值标量函数

$$f(\mathbf{X}),{\mathbf{X}}_{m\times n}=(x_{ij}{)}_{i=1,j=1}^{m,n}$$利用梯度矩阵的形式，也就是矩阵变元的标量函数里的分母布局的形式，有： ∇ X f ( X ) = ∂ f ( X ) ∂ X m × n = [ ∂ f ∂ x 11 ∂ f ∂ x 12 ⋯ ∂ f ∂ x 1 n ∂ f ∂ x 21 ∂ f ∂ x 22 ⋯ ∂ f ∂ x 2 n ⋮ ⋮ ⋮ ⋮ ∂ f ∂ x m 1 ∂ f ∂ x m 2 ⋯ ∂ f ∂ x m n ] m × n \begin{aligned} \nabla_{X}f(\boldsymbol{X})& =\frac{\partial f(\boldsymbol{X})}{\partial\boldsymbol{X}_{m\times n}} \\ &=\begin{bmatrix}\frac{\partial f}{\partial x_{11}}&\frac{\partial f}{\partial x_{12}}&\cdots&\frac{\partial f}{\partial x_{1n}}\\\frac{\partial f}{\partial x_{21}}&\frac{\partial f}{\partial x_{22}}&\cdots&\frac{\partial f}{\partial x_{2n}}\\\vdots&\vdots&\vdots&\vdots\\\frac{\partial f}{\partial x_{m1}}&\frac{\partial f}{\partial x_{m2}}&\cdots&\frac{\partial f}{\partial x_{mn}}\end{bmatrix}_{m\times n} \end{aligned} ∇X​f(X)​=∂Xm×n​∂f(X)​= ​∂x11​∂f​∂x21​∂f​⋮∂xm1​∂f​​∂x12​∂f​∂x22​∂f​⋮∂xm2​∂f​​⋯⋯⋮⋯​∂x1n​∂f​∂x2n​∂f​⋮∂xmn​∂f​​ ​m×n​​类似于向量变元的实值标量函数，给出下面给个4个原则，并选择性给出证明：

2.1 4个法则

我们设讨论的矩阵 $X$ 是 $m\times n$ 维的

（1）：常数求导

与一元函数常数求导相同：结果为零矩阵，即

$$\frac{\partial c}{\partial X}=0_{m\times n}$$其中，$c$ 为常数

（2）：线性法则

与一元函数求导线性法则相同：相加再求导等于求导再相加，常数提到外面，即：$$\frac{\partial [c_{1}f(X)+c_{2}g(X)]}{\partial X}=c_1\frac{\partial f(X)}{\partial X}+c_2\frac{\partial g(X)}{\partial X}$$其中，$c_1,c_2$ 为常数。

（3）：乘积法则

与一元函数求导乘积法则相同：前导后不导加前不导后导，即$$\frac{\partial [f(\mathbf{X})g(\mathbf{X})]}{\partial \mathbf{X}}=\frac{\partial f(\mathbf{X})}{\partial \mathbf{X}}g(\mathbf{X})+f(\mathbf{X})\frac{\partial g(\mathbf{X})}{\partial \mathbf{X}}$$

证明：由于矩阵变元的实值标量函数是对逐一每个元素 $d{x}_{ij}$ 的导数，为此利用向量变元的实值标量函数的乘积法则有： ∂ [ f ( X ) g ( X ) ] ∂ X = [ ∂ ( f g ) ∂ x 11 ∂ ( f g ) ∂ x 12 ⋯ ∂ ( f g ) ∂ x 1 n ∂ ( f g ) ∂ x 21 ∂ ( f g ) ∂ x 22 ⋯ ∂ ( f g ) ∂ x 2 n ⋮ ⋮ ⋮ ⋮ ∂ ( f g ) ∂ x m 1 ∂ ( f g ) ∂ x m 2 ⋯ ∂ ( f g ) ∂ x m n ] = [ ∂ f ∂ x 11 g + f ∂ g ∂ x 11 ∂ f ∂ x 12 g + f ∂ g ∂ x 12 ⋯ ∂ f ∂ x 1 n g + f ∂ g ∂ x 1 n ∂ f ∂ x 21 g + f ∂ g ∂ x 21 ∂ f ∂ x 22 g + f ∂ g ∂ x 22 ⋯ ∂ f ∂ x 2 n g + f ∂ g ∂ x 2 n ⋮ ⋮ ⋮ ⋮ ∂ f ∂ x m 1 g + f ∂ g ∂ x m 1 ∂ f ∂ x m 2 g + f ∂ g ∂ x m 2 ⋯ ∂ f ∂ x m n g + f ∂ g ∂ x m n ] = [ ∂ f ∂ x 11 ∂ f ∂ x 12 ⋯ ∂ f ∂ x 1 n ∂ f ∂ x 21 ∂ f ∂ x 22 ⋯ ∂ f ∂ x 2 n ⋮ ⋮ ⋮ ⋮ ∂ f ∂ x m 1 ∂ f ∂ x m 2 ⋯ ∂ f ∂ x m n ] g + f [ ∂ g ∂ x 11 ∂ g ∂ x 12 ⋯ ∂ g ∂ x 1 n ∂ g ∂ x 21 ∂ g ∂ x 22 ⋯ ∂ g ∂ x 2 n ⋮ ⋮ ⋮ ⋮ ∂ g ∂ x m 1 ∂ g ∂ x m 2 ⋯ ∂ g ∂ x m n ] = ∂ f ( X ) ∂ X g ( X ) + f ( X ) ∂ g ( X ) ∂ X (23) \begin{aligned} \frac{\partial{[f(\pmb{X})g(\pmb{X})]}}{\partial{\pmb{X}}} &= \begin{bmatrix} \frac{\partial{(fg)}}{\partial{x_{11}}} & \frac{\partial{(fg)}}{\partial{x_{12}}} & \cdots & \frac{\partial{(fg)}}{\partial{x_{1n}}} \\ \frac{\partial{(fg)}}{\partial{x_{21}}} & \frac{\partial{(fg)}}{\partial{x_{22}}} & \cdots & \frac{\partial{(fg)}}{\partial{x_{2n}}} \\ \vdots & \vdots & \vdots & \vdots \\ \frac{\partial{(fg)}}{\partial{x_{m1}}} & \frac{\partial{(fg)}}{\partial{x_{m2}}} & \cdots & \frac{\partial{(fg)}}{\partial{x_{mn}}} \end{bmatrix} \\\\ &= \begin{bmatrix} \frac{\partial{f}}{\partial{x_{11}}}g+f\frac{\partial{g}}{\partial{x_{11}}} & \frac{\partial{f}}{\partial{x_{12}}}g+f\frac{\partial{g}}{\partial{x_{12}}} & \cdots & \frac{\partial{f}}{\partial{x_{1n}}}g+f\frac{\partial{g}}{\partial{x_{1n}}} \\ \frac{\partial{f}}{\partial{x_{21}}}g+f\frac{\partial{g}}{\partial{x_{21}}} & \frac{\partial{f}}{\partial{x_{22}}}g+f\frac{\partial{g}}{\partial{x_{22}}} & \cdots & \frac{\partial{f}}{\partial{x_{2n}}}g+f\frac{\partial{g}}{\partial{x_{2n}}}\\ \vdots & \vdots & \vdots & \vdots \\ \frac{\partial{f}}{\partial{x_{m1}}}g+f\frac{\partial{g}}{\partial{x_{m1}}} & \frac{\partial{f}}{\partial{x_{m2}}}g+f\frac{\partial{g}}{\partial{x_{m2}}} & \cdots & \frac{\partial{f}}{\partial{x_{mn}}}g+f\frac{\partial{g}}{\partial{x_{mn}}} \end{bmatrix} \\\\ &=\begin{bmatrix} \frac{\partial{f}}{\partial{x_{11}}}&\frac{\partial{f}}{\partial{x_{12}}}&\cdots&\frac{\partial{f}}{\partial{x_{1n}}} \\ \frac{\partial{f}}{\partial{x_{21}}}&\frac{\partial{f}}{\partial{x_{22}}}&\cdots&\frac{\partial{f}}{\partial{x_{2n}}} \\ \vdots &\vdots & \vdots & \vdots\\ \frac{\partial{f}}{\partial{x_{m1}}}&\frac{\partial{f}}{\partial{x_{m2}}}&\cdots&\frac{\partial{f}}{\partial{x_{mn}}} \end{bmatrix}g + f\begin{bmatrix}\frac{\partial{g}}{\partial{x_{11}}}&\frac{\partial{g}}{\partial{x_{12}}}&\cdots&\frac{\partial{g}}{\partial{x_{1n}}} \\ \frac{\partial{g}}{\partial{x_{21}}}&\frac{\partial{g}}{\partial{x_{22}}}&\cdots&\frac{\partial{g}}{\partial{x_{2n}}} \\ \vdots &\vdots & \vdots & \vdots\\ \frac{\partial{g}}{\partial{x_{m1}}}&\frac{\partial{g}}{\partial{x_{m2}}}&\cdots&\frac{\partial{g}}{\partial{x_{mn}}} \end{bmatrix} \\\\ &=\frac{\partial f(\pmb{X})}{\partial{\pmb{X}}}g(\pmb{X}) +f(\pmb{X})\frac{\partial g(\pmb{X})}{\partial{\pmb{X}}} \end{aligned} \\\\ \tag{23} ∂X∂[f(X)g(X)]​​= ​∂x11​∂(fg)​∂x21​∂(fg)​⋮∂xm1​∂(fg)​​∂x12​∂(fg)​∂x22​∂(fg)​⋮∂xm2​∂(fg)​​⋯⋯⋮⋯​∂x1n​∂(fg)​∂x2n​∂(fg)​⋮∂xmn​∂(fg)​​ ​= ​∂x11​∂f​g+f∂x11​∂g​∂x21​∂f​g+f∂x21​∂g​⋮∂xm1​∂f​g+f∂xm1​∂g​​∂x12​∂f​g+f∂x12​∂g​∂x22​∂f​g+f∂x22​∂g​⋮∂xm2​∂f​g+f∂xm2​∂g​​⋯⋯⋮⋯​∂x1n​∂f​g+f∂x1n​∂g​∂x2n​∂f​g+f∂x2n​∂g​