Matrix derivative w.r.t time $\frac{d}{dt}\|A-X(t)\|^2=2\operatorname{tr}\big((X-A)\dot{X}\big)$ My question is just how to derive this equality?   
$$\frac{d}{dt}\|A-X(t)\|^2=2\operatorname{tr}\big((X-A)\dot{X}\big)$$
where  


*

*$A, X\in \mathbb{R}^{n\times n}$   


In particular, how to obtain the matrices trace product term?
It seems there are three levels chain rule, one for square, one for norm and then $X(t)$
I just know the following:    $$\nabla_x \|Ax-b\|^2=2A^T(Ax-b)$$
 A: Let's use the $L^2$ norm as an example.  Then, we have
$$\begin{align}
||A-X(t)||^2&=\sum_{i,j}|A_{ij}-X_{ij}(t)|^2\\\\
&=\sum_{i,j}(A^2_{ij}-2A_{ij}X_{ij}(t)+X^2_{ij}(t))
\end{align}$$
Differentiating $(1)$ we obtain
$$\begin{align}
\frac{d}{dt}||A-X(t)||^2&=\sum_{i,j}(-2A_{ij}X'_{ij}(t)+2X_{ij}X_{ij}'(t))\\\\
&=2\sum_{ij}(X_{ij}(t)-A_{ij})X'_{ij}(t)\\\\
&=2\sum_{ij}(X_{ij}(t)-A_{ij})(X'_{ji})^T(t)\\\\
&=2\,\text{tr}\left((X(t)-A)X'(t)\right)
\end{align}$$
as was to be shown!

Note:  As elucidated in the comment by @mattos, the result can be obtained using inner product notatation.  We could have begun by writing $||A-X(t)||^2=\langle A-X(t),A-X(t)\rangle$.  Upon differentiating, we obtain $$\frac{d}{dt}||A-X(t)||^2=\langle X(t)-A,X'(t)\rangle +\langle X'(t),X(t)-A\rangle=2\text{tr}\left(A-X(t),X'(t)\right)$$as expected!

A: For a moment, consider the case where $X,\dot{X},A\in{\mathbb R}^{m\times n}$
Then your assumed answer cannot possibly be correct since the product 
$$(X-A)\dot{X}$$
 has incompatible dimensions.
Several posters have suggested working with the inner product (which I'll denote by a colon). If you follow that suggestion you obtain 
$$\eqalign{
 f &= (X-A):(X-A) \cr\cr
df &= 2(X-A):dX \cr
   &= 2(X-A):\dot{X} dt \cr\cr
\frac{df}{dt} &= 2(X-A):\dot{X} \cr
 &= 2\operatorname{tr}\big((X-A)^T\dot{X}\big) \cr\cr
}$$
This result is correct for the rectangular $(m\times n)$ case and also when $(m=n)$.
If you're sure that your assumed answer is correct, then perhaps you forgot to mention that $X$ is symmetric?
