Proving a rule about del operator as applied to matrices

Question

How can I prove the following easily?(If it is true of course.)

\begin{align} \nabla_{\mathbf{x}_k} \left( \sum_{i=1}^{n}\sum_{j=1}^{n} \mathbf{x}^{T}_i \mathbf{W}_{ij} \mathbf{x}_j \right)=\sum_{j=1}^{n}(\mathbf{W}_{kj}\mathbf{x}_{j}+\mathbf{W}_{jk}^{T}\mathbf{x}_j) \end{align}

Here each $\mathbf{x}_i \in \mathbb{R}^{N}$ is a different vector, and each $\mathbf{W}_{ij} \in \mathbb{R}^{N\times N}$ is a different matrix of real numbers for $i,j = 1,2,...,n$.

Answer 1

I think I solved: \begin{align} \nabla_{x_k} \left( \sum_{i=1}^{n}\sum_{j=1}^{n} x^{T}_i W_{ij} x_j \right) &= \sum_{i=1,i\neq k}^{n}W_{kj}x_j + \sum_{j=1, j\neq k}^{n}(x^T_jW_{jk})^T + x^T_k W_{kk} x_k\\ \text{using the linked question:}\\ &= \sum_{i=1,i\neq k}^{n}\left( W_{ki}x_i + W^T_{ik}x_i \right)+ W_{kk} x_k + W^T_{kk}x_k\\ &= \sum_{i=1}^{n}\left( W_{ki}x_i + W^T_{ik}x_i \right) \end{align}