TITLE: Derivative of Quadratic Form
QUESTION [31 upvotes]: For the Quadratic Form $X^TAX; X\in\mathbb{R}^n, A\in\mathbb{R}^{n \times n}$ (which simplifies to $\Sigma_{i=0}^n\Sigma_{j=0}^nA_{ij}x_ix_j$), I tried to take the derivative wrt. X ($\Delta_X X^TAX$) and ended up with the following:
The $k^{th}$ element of the derivative represented as
$\Delta_{X_k}X^TAX=[\Sigma_{i=1}^n(A_{ik}x_k+A_{ki})x_i] + A_{kk}x_k(1-x_k)$
Does this result look right? Is there an alternative form?
I'm trying to get to the $\mu_0$ of Gaussian Discriminant Analysis by maximizing the log likelihood and I need to take the derivative of a Quadratic form. Either the result I mentioned above is wrong (shouldn't be because I went over my arithmetic several times) or the form I arrived at above is not the terribly useful to my problem (because I'm unable to proceed).
I can give more details about the problem or the steps I put down to arrive at the above result, but I didn't want to clutter to start off. Please let me know if more details are necessary.
Any link to related material is also much appreciated.

REPLY [4 votes]: Yet another approach using the Frobenius product notation.
For a column vector $x \in \mathbb{R}^n$, and a matrix $A \in \mathbb{R}^{n \times n}$  we can write:
$$ x^TAx = Tr(x^TAx) = x:Ax$$
Then we take the differential and derivative as
\begin{align}
    d(x:Ax) & = dx:Ax + x:Adx\\
            & = Ax:dx + A^Tx:dx\\
            & = (Ax + A^Tx):dx\\
         \frac{\partial (x^TAx)}{\partial x} &= (Ax + A^Tx) = (A + A^T)x 
\end{align}