Derivative of matrix product: is it true that $\frac{d}{dt}(A^TA) = 2A^T \frac{dA}{dt}$? $A$ is a square matrix.  All elements of $A$ depend on a parameter $t$, that is, $a_{ij}=a_{ij}(t)$.  Let $S(A):=A^TA$, and take the derivative of $S$ w.r.t. $t$:
$\displaystyle \frac{dS}{dt}$
Now, pretty clearly
$\displaystyle \frac{dS}{dt} = \frac{dA^T}{dt}A + A^T\frac{dA}{dt}$
But now, can this also be written $\displaystyle \frac{dS}{dt} = 2 A^T \frac{dA}{dt}$  ?
A math text I am working though right now -- if I am reading it right -- implies that this is the case, but I haven't been able to prove it myself.
Thanks.

Even when one assume $A(t)$ is symmetric, the alternate form is still incorrect.
For a counter-example, consider
$$A(t) = \begin{bmatrix}0 & 1 & 0\\1 & 0 & t\\0 & t & 0\end{bmatrix}$$
We have 
$$\frac{d}{dt} S(t) = 
\begin{bmatrix}0 & 0 & 1\\0 & 2t & 0\\1 & 0 & 2t\end{bmatrix}
\ne
\begin{bmatrix}0 & 0 & 2\\0 & 2t & 0\\0 & 0 & 2t\end{bmatrix}
= 2A(t)\frac{d}{dt}A(t)$$
