矩阵二范数的求导问题

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
$$\begin{array}{rl}\frac{\partial f(x)}{\partial x}& =\frac{1}{2}\{[A^{T}A+(A^{T}A{)}^T]x-A^{T}b-A^{T}b\}\\ & =\frac{1}{2}\{2A^{T}Ax-2A^{T}b\}\\ & =A^{T}Ax-2A^{T}b\\ & =A^T(Ax-b)\end{array}$$

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