Derivative of quadratic form 3 Could someone provide me with step by step derivation of derivative of quadratic form:
$\alpha=\sum_{i=1}^{m}\sum_{j=1}^{m}a_{ij}x_ix_j $ ?
Specifically, I am interested in obtaining derivative in such form : $\frac{d\alpha}{dx_k}=\sum_{j=1}^{m}a_{kj}x_j+\sum_{i=1}^{n}a_{ki}x_i$ which is $\frac{d\alpha}{dx}=\textbf{x}^T(A^T+A)$ which is simply $2x^TA$ when A is symmetric matrix.
So another detailed justification is the following. Expand $\alpha = x^TAx$ as you did, namely:
$$\alpha  = x^TAx = \sum_{i=1}^{m}\sum_{j=1}^{m}a_{ij}x_ix_j $$
Derive w.r.t the $k^{th}$ element of $x$ (say $x_k$), you get
$$\frac{\partial \alpha}{\partial x_k} = 2a_{kk}x_k + \sum\limits_{i=1 \\ i\neq k}^m (a_{ik} + a_{ki})x_i = \sum\limits_{i=1}^m (a_{ik} + a_{ki})x_i \qquad k = 1\ldots m$$
So $$\frac{\partial \alpha}{\partial x}=
  \begin{bmatrix}
    \frac{\partial \alpha}{\partial x_1} \\
    \frac{\partial \alpha}{\partial x_2} \\
\vdots \\
\frac{\partial \alpha}{\partial x_m} 
  \end{bmatrix}
=
 \begin{bmatrix}
    \sum\limits_{i=1}^m (a_{i1} + a_{1i})x_i \\
    \sum\limits_{i=1}^m (a_{i2} + a_{2i})x_i \\
\vdots \\
\sum\limits_{i=1}^m (a_{im} + a_{mi})x_i
  \end{bmatrix}
=x^T(A + A^T)
$$
So $\alpha = x^T A x$. 

Using $\frac{d x^T A}{d x} = A$, then $\frac{d x^T Ax}{d x} = Ax + A^Tx$