[足式机器人]Part2 Dr. CAN学习笔记-Ch0-1矩阵的导数运算

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1. 标量向量方程对向量求导，分母布局，分子布局

1.1 标量方程对向量的导数

• y y y 为 一元向量 或 二元向量
  
• y y y 为多元向量
  y ⃗ = [ y 1 , y 2 , ⋯   , y n ] ⇒ ∂ f ( y ⃗ ) ∂ y ⃗ \vec{y}=\left[ y_1,y_2,\cdots ,y_{\mathrm{n}} \right] \Rightarrow \frac{\partial f\left( \vec{y} \right)}{\partial \vec{y}} y ​ = [ y 1 ​ , y 2 ​ , ⋯ , y n ​ ] ⇒ ∂ y ​ ∂ f ( y ​ ) ​
  其中： f ( y ⃗ ) f\left( \vec{y} \right) f ( y ​ ) 为标量 1 × 1 1\times 1 1 × 1 , y ⃗ \vec{y} y ​ 为向量 1 × n 1\times n 1 × n

1. 分母布局 Denominator Layout ——行数与分母相同
   ∂ f ( y ⃗ ) ∂ y ⃗ = [ ∂ f ( y ⃗ ) ∂ y 1 ⋮ ∂ f ( y ⃗ ) ∂ y n ] n × 1 \frac{\partial f\left( \vec{y} \right)}{\partial \vec{y}}=\left[ \begin{array}{c} \frac{\partial f\left( \vec{y} \right)}{\partial y_1}\\ \vdots\\ \frac{\partial f\left( \vec{y} \right)}{\partial y_{\mathrm{n}}}\\ \end{array} \right] _{n\times 1} ∂ y ​ ∂ f ( y ​ ) ​ = ​ ∂ y 1 ​ ∂ f ( y ​ ) ​ ⋮ ∂ y n ​ ∂ f ( y ​ ) ​ ​ ​ n × 1 ​
2. 分子布局 Nunerator Layout ——行数与分子相同
   ∂ f ( y ⃗ ) ∂ y ⃗ = [ ∂ f ( y ⃗ ) ∂ y 1 ⋯ ∂ f ( y ⃗ ) ∂ y n ] 1 × n \frac{\partial f\left( \vec{y} \right)}{\partial \vec{y}}=\left[ \begin{matrix} \frac{\partial f\left( \vec{y} \right)}{\partial y_1}& \cdots& \frac{\partial f\left( \vec{y} \right)}{\partial y_{\mathrm{n}}}\\ \end{matrix} \right] _{1\times n} ∂ y ​ ∂ f ( y ​ ) ​ = [ ∂ y 1 ​ ∂ f ( y ​ ) ​ ​ ⋯ ​ ∂ y n ​ ∂ f ( y ​ ) ​ ​ ] 1 × n ​

1.2 向量方程对向量的导数

f ⃗ ( y ⃗ ) = [ f ⃗ 1 ( y ⃗ ) ⋮ f ⃗ n ( y ⃗ ) ] n × 1 , y ⃗ = [ y 1 ⋮ y m ] m × 1 \vec{f}\left( \vec{y} \right) =\left[ \begin{array}{c} \vec{f}_1\left( \vec{y} \right)\\ \vdots\\ \vec{f}_{\mathrm{n}}\left( \vec{y} \right)\\ \end{array} \right] _{n\times 1},\vec{y}=\left[ \begin{array}{c} y_1\\ \vdots\\ y_{\mathrm{m}}\\ \end{array} \right] _{\mathrm{m}\times 1} f ​ ( y ​ ) = ​ f ​ 1 ​ ( y ​ ) ⋮ f ​ n ​ ( y ​ ) ​ ​ n × 1 ​ , y ​ = ​ y 1 ​ ⋮ y m ​ ​ ​ m × 1 ​
∂ f ⃗ ( y ⃗ ) n × 1 ∂ y ⃗ m × 1 = [ ∂ f ⃗ ( y ⃗ ) ∂ y 1 ⋮ ∂ f ⃗ ( y ⃗ ) ∂ y m ] m × 1 = [ ∂ f 1 ( y ⃗ ) ∂ y 1 ⋯ ∂ f n ( y ⃗ ) ∂ y 1 ⋮ ⋱ ⋮ ∂ f 1 ( y ⃗ ) ∂ y m ⋯ ∂ f n ( y ⃗ ) ∂ y m ] m × n \frac{\partial \vec{f}\left( \vec{y} \right) _{n\times 1}}{\partial \vec{y}_{\mathrm{m}\times 1}}=\left[ \begin{array}{c} \frac{\partial \vec{f}\left( \vec{y} \right)}{\partial y_1}\\ \vdots\\ \frac{\partial \vec{f}\left( \vec{y} \right)}{\partial y_{\mathrm{m}}}\\ \end{array} \right] _{\mathrm{m}\times 1}=\left[ \begin{matrix} \frac{\partial f_1\left( \vec{y} \right)}{\partial y_1}& \cdots& \frac{\partial f_{\mathrm{n}}\left( \vec{y} \right)}{\partial y_1}\\ \vdots& \ddots& \vdots\\ \frac{\partial f_1\left( \vec{y} \right)}{\partial y_{\mathrm{m}}}& \cdots& \frac{\partial f_{\mathrm{n}}\left( \vec{y} \right)}{\partial y_{\mathrm{m}}}\\ \end{matrix} \right] _{\mathrm{m}\times \mathrm{n}} ∂ y ​ m × 1 ​ ∂ f ​ ( y ​ ) n × 1 ​ ​ = ​ ∂ y 1 ​ ∂ f ​ ( y ​ ) ​ ⋮ ∂ y m ​ ∂ f ​ ( y ​ ) ​ ​ ​ m × 1 ​ = ​ ∂ y 1 ​ ∂ f 1 ​ ( y ​ ) ​ ⋮ ∂ y m ​ ∂ f 1 ​ ( y ​ ) ​ ​ ⋯ ⋱ ⋯ ​ ∂ y 1 ​ ∂ f n ​ ( y ​ ) ​ ⋮ ∂ y m ​ ∂ f n ​ ( y ​ ) ​ ​ ​ m × n ​ , 为分母布局

若： y ⃗ = [ y 1 ⋮ y m ] m × 1 , A = [ a 11 ⋯ a 1 n ⋮ ⋱ ⋮ a m 1 ⋯ a m n ] \vec{y}=\left[ \begin{array}{c} y_1\\ \vdots\\ y_{\mathrm{m}}\\ \end{array} \right] _{\mathrm{m}\times 1}, A=\left[ \begin{matrix} a_{11}& \cdots& a_{1\mathrm{n}}\\ \vdots& \ddots& \vdots\\ a_{\mathrm{m}1}& \cdots& a_{\mathrm{mn}}\\ \end{matrix} \right] y ​ = ​ y 1 ​ ⋮ y m ​ ​ ​ m × 1 ​ , A = ​ a 11 ​ ⋮ a m 1 ​ ​ ⋯ ⋱ ⋯ ​ a 1 n ​ ⋮ a mn ​ ​ ​ , 则有：

• ∂ A y ⃗ ∂ y ⃗ = A T \frac{\partial A\vec{y}}{\partial \vec{y}}=A^{\mathrm{T}} ∂ y ​ ∂ A y ​ ​ = A T (分母布局)
• ∂ y ⃗ T A y ⃗ ∂ y ⃗ = A y ⃗ + A T y ⃗ \frac{\partial \vec{y}^{\mathrm{T}}A\vec{y}}{\partial \vec{y}}=A\vec{y}+A^{\mathrm{T}}\vec{y} ∂ y ​ ∂ y ​ T A y ​ ​ = A y ​ + A T y ​ , 当 A = A T A=A^{\mathrm{T}} A = A T 时, ∂ y ⃗ T A y ⃗ ∂ y ⃗ = 2 A y ⃗ \frac{\partial \vec{y}^{\mathrm{T}}A\vec{y}}{\partial \vec{y}}=2A\vec{y} ∂ y ​ ∂ y ​ T A y ​ ​ = 2 A y ​

若为分子布局，则有： ∂ A y ⃗ ∂ y ⃗ = A \frac{\partial A\vec{y}}{\partial \vec{y}}=A ∂ y ​ ∂ A y ​ ​ = A

2. 案例分析，线性回归

• ∂ A y ⃗ ∂ y ⃗ = A T \frac{\partial A\vec{y}}{\partial \vec{y}}=A^{\mathrm{T}} ∂ y ​ ∂ A y ​ ​ = A T (分母布局)
• ∂ y ⃗ T A y ⃗ ∂ y ⃗ = A y ⃗ + A T y ⃗ \frac{\partial \vec{y}^{\mathrm{T}}A\vec{y}}{\partial \vec{y}}=A\vec{y}+A^{\mathrm{T}}\vec{y} ∂ y ​ ∂ y ​ T A y ​ ​ = A y ​ + A T y ​ , 当 A = A T A=A^{\mathrm{T}} A = A T 时, ∂ y ⃗ T A y ⃗ ∂ y ⃗ = 2 A y ⃗ \frac{\partial \vec{y}^{\mathrm{T}}A\vec{y}}{\partial \vec{y}}=2A\vec{y} ∂ y ​ ∂ y ​ T A y ​ ​ = 2 A y ​

Linear Regression 线性回归
z ^ = y 1 + y 2 x ⇒ J = ∑ i = 1 n [ z i − ( y 1 + y 2 x i ) ] 2 \hat{z}=y_1+y_2x\Rightarrow J=\sum_{i=1}^n{\left[ z_i-\left( y_1+y_2x_i \right) \right] ^2} z ^ = y 1 ​ + y 2 ​ x ⇒ J = i = 1 ∑ n ​ [ z i ​ − ( y 1 ​ + y 2 ​ x i ​ ) ] 2
找到 y 1 , y 2 y_1,y_2 y 1 ​ , y 2 ​ 使得 J J J 最小

z ⃗ = [ z 1 ⋮ z n ] , [ x ⃗ ] = [ 1 x 1 ⋮ ⋮ 1 x n ] , y ⃗ = [ y 1 y 2 ] ⇒ z ⃗ ^ = [ x ⃗ ] y ⃗ = [ y 1 + y 2 x 1 ⋮ y 1 + y 2 x n ] \vec{z}=\left[ \begin{array}{c} z_1\\ \vdots\\ z_{\mathrm{n}}\\ \end{array} \right] ,\left[ \vec{x} \right] =\left[ \begin{array}{l} 1& x_1\\ \vdots& \vdots\\ 1& x_{\mathrm{n}}\\ \end{array} \right] ,\vec{y}=\left[ \begin{array}{c} y_1\\ y_2\\ \end{array} \right] \Rightarrow \hat{\vec{z}}=\left[ \vec{x} \right] \vec{y}=\left[ \begin{array}{c} y_1+y_2x_1\\ \vdots\\ y_1+y_2x_{\mathrm{n}}\\ \end{array} \right] z = ​ z 1 ​ ⋮ z n ​ ​ ​ , [ x ] = ​ 1 ⋮ 1 ​ x 1 ​ ⋮ x n ​ ​ ​ , y ​ = [ y 1 ​ y 2 ​ ​ ] ⇒ z ^ = [ x ] y ​ = ​ y 1 ​ + y 2 ​ x 1 ​ ⋮ y 1 ​ + y 2 ​ x n ​ ​ ​
J = [ z ⃗ − z ⃗ ^ ] T [ z ⃗ − z ⃗ ^ ] = [ z ⃗ − [ x ⃗ ] y ⃗ ] T [ z ⃗ − [ x ⃗ ] y ⃗ ] = z ⃗ z ⃗ T − z ⃗ T [ x ⃗ ] y ⃗ − y ⃗ T [ x ⃗ ] T z ⃗ + y ⃗ T [ x ⃗ ] T [ x ⃗ ] y ⃗ J=\left[ \vec{z}-\hat{\vec{z}} \right] ^{\mathrm{T}}\left[ \vec{z}-\hat{\vec{z}} \right] =\left[ \vec{z}-\left[ \vec{x} \right] \vec{y} \right] ^{\mathrm{T}}\left[ \vec{z}-\left[ \vec{x} \right] \vec{y} \right] =\vec{z}\vec{z}^{\mathrm{T}}-\vec{z}^{\mathrm{T}}\left[ \vec{x} \right] \vec{y}-\vec{y}^{\mathrm{T}}\left[ \vec{x} \right] ^{\mathrm{T}}\vec{z}+\vec{y}^{\mathrm{T}}\left[ \vec{x} \right] ^{\mathrm{T}}\left[ \vec{x} \right] \vec{y} J = [ z − z ^ ] T [ z − z ^ ] = [ z − [ x ] y ​ ] T [ z − [ x ] y ​ ] = z z T − z T [ x ] y ​ − y ​ T [ x ] T z + y ​ T [ x ] T [ x ] y ​
其中： ( z ⃗ T [ x ⃗ ] y ⃗ ) T = y ⃗ T [ x ⃗ ] T z ⃗ \left( \vec{z}^{\mathrm{T}}\left[ \vec{x} \right] \vec{y} \right) ^{\mathrm{T}}=\vec{y}^{\mathrm{T}}\left[ \vec{x} \right] ^{\mathrm{T}}\vec{z} ( z T [ x ] y ​ ) T = y ​ T [ x ] T z ， 则有：
J = z ⃗ z ⃗ T − 2 z ⃗ T [ x ⃗ ] y ⃗ + y ⃗ T [ x ⃗ ] T [ x ⃗ ] y ⃗ J=\vec{z}\vec{z}^{\mathrm{T}}-2\vec{z}^{\mathrm{T}}\left[ \vec{x} \right] \vec{y}+\vec{y}^{\mathrm{T}}\left[ \vec{x} \right] ^{\mathrm{T}}\left[ \vec{x} \right] \vec{y} J = z z T − 2 z T [ x ] y ​ + y ​ T [ x ] T [ x ] y ​
进而：
∂ J ∂ y ⃗ = 0 − 2 ( z ⃗ T [ x ⃗ ] ) T + 2 [ x ⃗ ] T [ x ⃗ ] y ⃗ = ∇ y ⃗ ⟹ ∂ J ∂ y ⃗ ∗ = 0 , y ⃗ ∗ = ( [ x ⃗ ] T [ x ⃗ ] ) − 1 [ x ⃗ ] T z ⃗ \frac{\partial J}{\partial \vec{y}}=0-2\left( \vec{z}^{\mathrm{T}}\left[ \vec{x} \right] \right) ^{\mathrm{T}}+2\left[ \vec{x} \right] ^{\mathrm{T}}\left[ \vec{x} \right] \vec{y}=\nabla \vec{y}\Longrightarrow \frac{\partial J}{\partial \vec{y}^*}=0,\vec{y}^*=\left( \left[ \vec{x} \right] ^{\mathrm{T}}\left[ \vec{x} \right] \right) ^{-1}\left[ \vec{x} \right] ^{\mathrm{T}}\vec{z} ∂ y ​ ∂ J ​ = 0 − 2 ( z T [ x ] ) T + 2 [ x ] T [ x ] y ​ = ∇ y ​ ⟹ ∂ y ​ ∗ ∂ J ​ = 0 , y ​ ∗ = ( [ x ] T [ x ] ) − 1 [ x ] T z
其中： ( [ x ⃗ ] T [ x ⃗ ] ) − 1 \left( \left[ \vec{x} \right] ^{\mathrm{T}}\left[ \vec{x} \right] \right) ^{-1} ( [ x ] T [ x ] ) − 1 不一定有解，则 y ⃗ ∗ \vec{y}^* y ​ ∗ 无法得到解析解——定义初始 y ⃗ ∗ \vec{y}^* y ​ ∗ ， y ⃗ ∗ = y ⃗ ∗ − α ∇ , α = [ α 1 0 0 α 2 ] \vec{y}^*=\vec{y}^*-\alpha \nabla ,\alpha =\left[ \begin{matrix} \alpha _1& 0\\ 0& \alpha _2\\ \end{matrix} \right] y ​ ∗ = y ​ ∗ − α ∇ , α = [ α 1 ​ 0 ​ 0 α 2 ​ ​ ]
其中： α \alpha α 称为学习率，对 x x x 而言则需进行归一化

3. 矩阵求导的链式法则

标量函数： J = f ( y ( u ) ) , ∂ J ∂ u = ∂ J ∂ y ∂ y ∂ u J=f\left( y\left( u \right) \right) ,\frac{\partial J}{\partial u}=\frac{\partial J}{\partial y}\frac{\partial y}{\partial u} J = f ( y ( u ) ) , ∂ u ∂ J ​ = ∂ y ∂ J ​ ∂ u ∂ y ​

标量对向量求导： J = f ( y ⃗ ( u ⃗ ) ) , y ⃗ = [ y 1 ( u ⃗ ) ⋮ y m ( u ⃗ ) ] m × 1 , u ⃗ = [ u ⃗ 1 ⋮ u ⃗ n ] n × 1 J=f\left( \vec{y}\left( \vec{u} \right) \right) ,\vec{y}=\left[ \begin{array}{c} y_1\left( \vec{u} \right)\\ \vdots\\ y_{\mathrm{m}}\left( \vec{u} \right)\\ \end{array} \right] _{m\times 1},\vec{u}=\left[ \begin{array}{c} \vec{u}_1\\ \vdots\\ \vec{u}_{\mathrm{n}}\\ \end{array} \right] _{\mathrm{n}\times 1} J = f ( y ​ ( u ) ) , y ​ = ​ y 1 ​ ( u ) ⋮ y m ​ ( u ) ​ ​ m × 1 ​ , u = ​ u 1 ​ ⋮ u n ​ ​ ​ n × 1 ​

分析： ∂ J 1 × 1 ∂ u n × 1 n × 1 = ∂ J ∂ y m × 1 m × 1 ∂ y m × 1 ∂ u n × 1 n × m \frac{\partial J_{1\times 1}}{\partial u_{\mathrm{n}\times 1}}_{\mathrm{n}\times 1}=\frac{\partial J}{\partial y_{m\times 1}}_{m\times 1}\frac{\partial y_{m\times 1}}{\partial u_{\mathrm{n}\times 1}}_{\mathrm{n}\times \mathrm{m}} ∂ u n × 1 ​ ∂ J 1 × 1 ​ ​ n × 1 ​ = ∂ y m × 1 ​ ∂ J ​ m × 1 ​ ∂ u n × 1 ​ ∂ y m × 1 ​ ​ n × m ​ 无法相乘

y ⃗ = [ y 1 ( u ⃗ ) y 2 ( u ⃗ ) ] 2 × 1 , u ⃗ = [ u ⃗ 1 u ⃗ 2 u ⃗ 3 ] 3 × 1 \vec{y}=\left[ \begin{array}{c} y_1\left( \vec{u} \right)\\ y_2\left( \vec{u} \right)\\ \end{array} \right] _{2\times 1},\vec{u}=\left[ \begin{array}{c} \vec{u}_1\\ \vec{u}_2\\ \vec{u}_3\\ \end{array} \right] _{3\times 1} y ​ = [ y 1 ​ ( u ) y 2 ​ ( u ) ​ ] 2 × 1 ​ , u = ​ u 1 ​ u 2 ​ u 3 ​ ​ ​ 3 × 1 ​
J = f ( y ⃗ ( u ⃗ ) ) , ∂ J ∂ u ⃗ = [ ∂ J ∂ u ⃗ 1 ∂ J ∂ u ⃗ 2 ∂ J ∂ u ⃗ 3 ] 3 × 1 ⟹ ∂ J ∂ u ⃗ 1 = ∂ J ∂ y 1 ∂ y 1 ( u ⃗ ) ∂ u ⃗ 1 + ∂ J ∂ y 2 ∂ y 2 ( u ⃗ ) ∂ u ⃗ 1 ∂ J ∂ u ⃗ 2 = ∂ J ∂ y 1 ∂ y 1 ( u ⃗ ) ∂ u ⃗ 2 + ∂ J ∂ y 2 ∂ y 2 ( u ⃗ ) ∂ u ⃗ 2 ∂ J ∂ u ⃗ 3 = ∂ J ∂ y 1 ∂ y 1 ( u ⃗ ) ∂ u ⃗ 3 + ∂ J ∂ y 2 ∂ y 2 ( u ⃗ ) ∂ u ⃗ 3 ⟹ ∂ J ∂ u ⃗ = [ ∂ y 1 ( u ⃗ ) ∂ u ⃗ 1 ∂ y 2 ( u ⃗ ) ∂ u ⃗ 1 ∂ y 1 ( u ⃗ ) ∂ u ⃗ 2 ∂ y 2 ( u ⃗ ) ∂ u ⃗ 2 ∂ y 1 ( u ⃗ ) ∂ u ⃗ 3 ∂ y 2 ( u ⃗ ) ∂ u ⃗ 3 ] 3 × 2 [ ∂ J ∂ y 1 ∂ J ∂ y 2 ] 2 × 2 = ∂ y ⃗ ( u ⃗ ) ∂ u ⃗ ∂ J ∂ y ⃗ J=f\left( \vec{y}\left( \vec{u} \right) \right) ,\frac{\partial J}{\partial \vec{u}}=\left[ \begin{array}{c} \frac{\partial J}{\partial \vec{u}_1}\\ \frac{\partial J}{\partial \vec{u}_2}\\ \frac{\partial J}{\partial \vec{u}_3}\\ \end{array} \right] _{3\times 1}\Longrightarrow \begin{array}{c} \frac{\partial J}{\partial \vec{u}_1}=\frac{\partial J}{\partial y_1}\frac{\partial y_1\left( \vec{u} \right)}{\partial \vec{u}_1}+\frac{\partial J}{\partial y_2}\frac{\partial y_2\left( \vec{u} \right)}{\partial \vec{u}_1}\\ \frac{\partial J}{\partial \vec{u}_2}=\frac{\partial J}{\partial y_1}\frac{\partial y_1\left( \vec{u} \right)}{\partial \vec{u}_2}+\frac{\partial J}{\partial y_2}\frac{\partial y_2\left( \vec{u} \right)}{\partial \vec{u}_2}\\ \frac{\partial J}{\partial \vec{u}_3}=\frac{\partial J}{\partial y_1}\frac{\partial y_1\left( \vec{u} \right)}{\partial \vec{u}_3}+\frac{\partial J}{\partial y_2}\frac{\partial y_2\left( \vec{u} \right)}{\partial \vec{u}_3}\\ \end{array} \\ \Longrightarrow \frac{\partial J}{\partial \vec{u}}=\left[ \begin{array}{l} \frac{\partial y_1\left( \vec{u} \right)}{\partial \vec{u}_1}& \frac{\partial y_2\left( \vec{u} \right)}{\partial \vec{u}_1}\\ \frac{\partial y_1\left( \vec{u} \right)}{\partial \vec{u}_2}& \frac{\partial y_2\left( \vec{u} \right)}{\partial \vec{u}_2}\\ \frac{\partial y_1\left( \vec{u} \right)}{\partial \vec{u}_3}& \frac{\partial y_2\left( \vec{u} \right)}{\partial \vec{u}_3}\\ \end{array} \right] _{3\times 2}\left[ \begin{array}{c} \frac{\partial J}{\partial y_1}\\ \frac{\partial J}{\partial y_2}\\ \end{array} \right] _{2\times 2}=\frac{\partial \vec{y}\left( \vec{u} \right)}{\partial \vec{u}}\frac{\partial J}{\partial \vec{y}} J = f ( y ​ ( u ) ) , ∂ u ∂ J ​ = ​ ∂ u 1 ​ ∂ J ​ ∂ u 2 ​ ∂ J ​ ∂ u 3 ​ ∂ J ​ ​ ​ 3 × 1 ​ ⟹ ∂ u 1 ​ ∂ J ​ = ∂ y 1 ​ ∂ J ​ ∂ u 1 ​ ∂ y 1 ​ ( u ) ​ + ∂ y 2 ​ ∂ J ​ ∂ u 1 ​ ∂ y 2 ​ ( u ) ​ ∂ u 2 ​ ∂ J ​ = ∂ y 1 ​ ∂ J ​ ∂ u 2 ​ ∂ y 1 ​ ( u ) ​ + ∂ y 2 ​ ∂ J ​ ∂ u 2 ​ ∂ y 2 ​ ( u ) ​ ∂ u 3 ​ ∂ J ​ = ∂ y 1 ​ ∂ J ​ ∂ u 3 ​ ∂ y 1 ​ ( u ) ​ + ∂ y 2 ​ ∂ J ​ ∂ u 3 ​ ∂ y 2 ​ ( u ) ​ ​ ⟹ ∂ u ∂ J ​ = ​ ∂ u 1 ​ ∂ y 1 ​ ( u ) ​ ∂ u 2 ​ ∂ y 1 ​ ( u ) ​ ∂ u 3 ​ ∂ y 1 ​ ( u ) ​ ​ ∂ u 1 ​ ∂ y 2 ​ ( u ) ​ ∂ u 2 ​ ∂ y 2 ​ ( u ) ​ ∂ u 3 ​ ∂ y 2 ​ ( u ) ​ ​ ​ 3 × 2 ​ [ ∂ y 1 ​ ∂ J ​ ∂ y 2 ​ ∂ J ​ ​ ] 2 × 2 ​ = ∂ u ∂ y ​ ( u ) ​ ∂ y ​ ∂ J ​