Matrix derivative of $\operatorname{tr}\theta^TX^TX{\theta}$ This is a part of the derivation of the normal equation and I am struggling with this part.
I don't get how $\operatorname{tr}\theta^TX^TX{\theta}$ can become $2X^TX\theta$....
I know that the derivative of $\operatorname{tr}(ABA^TC)$ respect to $A$ is equal to $CAB + C^TAB^T$ and the lecturer seems that he wants me to use this to derive it, but I don't get how I should use it.
The picture is the part of the lecture note that I'm struggling with.

 A: Let's rewrite it using the trace/Frobenius product notation (colon), i.e.
\begin{align}
F & = \theta^TX^TX\theta\\
\implies Tr(F)&= Tr((X\theta)^TX\theta) = X \theta:X\theta\\
\implies dF & = d(X \theta):X\theta + X\theta:d(X \theta) \\
 & = 2X\theta:d(X \theta)\\
 & = 2X\theta: (d(X) \theta + X d\theta)\\
 & = 2X\theta: X d\theta\\
 & = 2X^TX\theta: d\theta\\
\implies \frac{dF}{d\theta} &=2X^TX\theta
\end{align}
=========================================
NB: I used the following properties of the trace function:
$$Tr(A^TB) = A:B$$
$$Tr(AB) = Tr(BA)$$
$$A:BC=AC^T:B=B^TA:C$$
A: \begin{align}
& \frac d {d\theta} \operatorname{tr}(\theta^TX^TX\theta) \\[12pt]
= {} & \frac d {dA}\operatorname{tr}(ABA^TC) \\[4pt]
& \text{with $\theta^T$ in the role of } A, \\
& \text{$X^TX$ in the role of } B, \\
& \text{and } I \text{ in the role of } C \\[12pt]
= {} & CAB + C^TAB^T \quad \text{(This was given.)} \\[10pt]
= {} & \theta^T X^TX + \theta^T X^TX.{}
\end{align}
Here, $B$ and $C$ must be square matrices, and their sizes differ if $A$ is not a square matrix.
