向量二范数的求导问题

现有目标函数：
$$f(x)=\frac{1}{2}\parallel Ax-b{\parallel }_2^2$$
其中$A\in {\mathbb{R}}^{N\times m}$,$x\in {\mathbb{R}}^m$,$b\in {\mathbb{R}}^n$
则求$f^{{}^{\prime }}(x)$的推导如下：
$$\begin{array}{rl}f(x)& =\frac{1}{2}\parallel Ax-b{\parallel }_2^2\\ & =\frac{1}{2}(Ax-b{)}^T(Ax-b)\\ & =\frac{1}{2}(x^T{A}^T-b^T)(Ax-b)\\ & =\frac{1}{2}(x^T{A}^{T}Ax-x^T{A}^{T}b-b^{T}Ax+b^{T}b)\end{array}$$
利用如下性质：
$$\begin{array}{rl}\frac{\partial x^{T}Dx}{\partial x}& =(D+D^T)x\\ \frac{\partial D^{T}x}{\partial x}& =D\\ \frac{\partial x^{T}D}{\partial x}& =D\end{array}$$
则有
∂f(x)∂x=12{[ATA+(ATA)T]x−ATb−ATb}=12{2ATAx−2ATb}=ATAx−2ATb=AT(Ax−b) \begin {aligned} \frac{\partial f(x)}{\partial x}&=\frac{1}{2} \{[A^TA+(A^TA)^T]x-A^Tb-A^Tb \} \\ &=\frac{1}{2} \{ 2A^TAx-2A^Tb \}\\ &=A^TAx-2A^Tb\\ &=A^T(Ax-b) \end {aligned} ∂x∂f(x)​​=21​{[ATA+(ATA)T]x−ATb−ATb}=21​{2ATAx−2ATb}=ATAx−2ATb=AT(Ax−b)​

故
$$\frac{\partial f(x)}{\partial x}=A^T(Ax-b)$$