Deriving a Subgradient for l2 norm composition over affine function

Question

I need to explain how to compute a subgradient at a given point x (not necessarily deriving it but just explain how it can be derived) for the following function: $$\newcommand{\defeq}{\mathrel{:\mkern-0.25mu=}}$$ $$f(\mathbf{x}) = \vert\vert \Sigma_{i=1}^{n} x_{i} \mathbf{A_{i}} -\mathbf{B} \vert \vert _{2}$$ where $\forall i \mathbf{A_{i}}, \mathbf{B} \in \mathbb{R}^{p\times q}$ and $\mathbf{x} \in \mathbb{R}^{n}$
My direction so far was to look at the definition of 2 norm of matrices $\vert\vert \mathbf{B} \vert \vert _{2} = \sqrt{\lambda_{max} (\mathbf{B}^{*}\mathbf{B}})$ and trying to use the fact that $A(\mathbf{x})^{*}A(\mathbf{x})$ is a convex function of $\mathbf{x}$ for $A(\mathbf{x}) \defeq \Sigma_{i=1}^{n} x_{i} \mathbf{A_{i}} -\mathbf{B} $ and therefor $\lambda_{max} (A(\mathbf{x})^{*}A(\mathbf{x}))$ has a subgradient (you can see that by the fact that $\lambda_{max}(A(\mathbf{x})^{*}A(\mathbf{x})) = max_{ \vert \vert \mathbf{u} \vert \vert =1} \mathbf{u}^{*} (\mathbf{A(\mathbf{x})^{*}A(\mathbf{x}))} \mathbf{u} $ is a maximum over convex functions of x).
The problem is that from there I am stuck, I want to use some sort of chain rule but $\sqrt{x}$ is not a convex function...
I am new to the subject of subgradients and I have been stuck on this exercise for a few days already, any help would be most welcomed.
Thanks in advance