if A,B are matrices, does the product rule hold like i've written it? (It holds if $\vec{x}$ was a scalar. ) $$\frac{d A(\vec{x})B(\vec{x})}{d\vec{x}} = \frac{d A(\vec{x})}{d\vec{x}} B(\vec{x}) + A(\vec{x})\frac{d B(\vec{x})}{d\vec{x}} $$
I do not know what the derivative $\frac{d A(\vec{x})}{d\vec{x}}$ is, so it is difficult to verify the above statement. But in my case, if the above identity holds, then there are eliminations and it is not required to calculate it.
Extra: Are there some sources which present the above in an not so formal manner? (I am an engineer)
Unfortunately, the proposed rule is false, but you can use index notation to derive the correct rule $$\eqalign{ \def\LR#1{\left(#1\right)} \def\p{\partial} \def\grad#1#2{\frac{d #1}{d #2}} \def\gradLR#1#2{\LR{\grad{#1}{#2}}} \def\op#1{\operatorname{#1}} \def\vecc#1{\op{vec}\LR{#1}} P &= AB \\ P_{ij} &= \sum_n A_{in}B_{nj} \\ \grad{P_{ij}}{x_k} &= \sum_n \gradLR{A_{in}}{x_k} B_{n j} + A_{in} \gradLR{B_{nj}}{x_k} \\ }$$ This result is a third-order tensor which cannot be written using standard matrix notation. The first term on the RHS is especially problematic.
One trick is to vectorize the equation before differentiating $$\eqalign{ p &= \vecc{P} \\ &= \LR{B\otimes I}^Ta &\doteq\; \LR{I\otimes A}b \\ \grad px &= \LR{B\otimes I}^T\gradLR ax \;&+\; \LR{I\otimes A}\gradLR bx \\ }$$ Vectorization flattens matrices into vectors, so matrix notation will work again.
To learn more about this subject, explore the matrix-calculus tag on this site.
The Matrix Cookbook is also a handy reference.