因变量为标量,自变量为向量
参考
\(y\) 为因变量,标量;\(X=[x_1,x_2,\dots,x_n]^T\) 为自变量是向量,n维。
\(y=f(X)\),即!!\(y = f(x_1,x_2,\dots,x_n)\)
因此可以直接求导:
求导结果为n维向量
以\(y = \vec a ^T\vec x\):表示y为两个向量的内积,结果为一个标量
则求\(\frac{\partial y}{\partial \vec x}\),只需求出所有的\(\frac{\partial y}{\partial \vec x_i}\)即可。
具体方法为:
将\(y\)的表达式展开成累加和的形式,然后套用标量的求导法则即可,这一方法适用于所有多维情况的求导。
解:
故对\(\forall i\):
\[\frac{\partial y}{\partial x_i} = a_i \]故:
\[\begin{aligned} \frac{\partial y}{\partial \vec x}&=(\frac{\partial y}{\partial x_1};\frac{\partial y}{\partial x_2};\dots;\frac{\partial y}{\partial x_n}) \\ ~&=(a_1;a_2;\dots ;a_n) \\ ~&=a \end{aligned} \]例子:
注意图中,向量\(x\)与\(w\)均写成了1n的形式,而不是我们通常的n1,因此最终算出来的结果里面为\(x^T\),而不是\(x\)
因变量、自变量均为向量
当自变量和因变量均为向量时,求导结果为一个矩阵,我们称该矩阵为雅可比矩阵(Jacobian Matrix)。
特别的,如果X为n*m的矩阵,w为m维向量,则
\[\frac{\partial X}{\partial \vec w} = X \]证明:
设
则,
\[\vec z=Xw=\begin{bmatrix} x_{11}w_1+x_{12}w_2+\dots+x_{1m}w_m\\ x_{21}w_1+x_{22}w_2+\dots+x_{2m}w_m\\ \vdots\\ x_{n1}w_1+x_{n2}w_2+\dots+x_{nm}w_m \end{bmatrix}=\begin{bmatrix} z_1\\ z_2\\ \vdots\\ z_n \end{bmatrix} \]则
\[\begin{aligned} \frac{\partial X\vec w}{\partial \vec w} &= \frac{\partial \vec z}{\partial \vec w}\\ &=\begin{bmatrix} \frac{\partial z_1}{\partial w_1}&\frac{\partial z_1}{\partial w_2}&\dots&\frac{\partial z_1}{\partial w_m}\\ \frac{\partial z_2}{\partial w_1}&\frac{\partial z_2}{\partial w_2}&\dots&\frac{\partial z_2}{\partial w_m}\\ \vdots&\vdots&\ddots&\vdots\\ \frac{\partial z_n}{\partial w_1}&\frac{\partial z_n}{\partial w_2}&\dots&\frac{\partial z_n}{\partial w_m}\\ \end{bmatrix}\\ &=\begin{bmatrix} x_{11}&x_{12}&\dots&x_{1m}\\ x_{21}&x_{22}&\dots&x_{2m}\\ \vdots&\vdots&\ddots&\vdots\\ x_{n1}&x_{n2}&\dots&x_{nm} \end{bmatrix}\\ &=X \end{aligned} \]例子: