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Let say that $A \in R^{n,m}$ and $x \in R^{m,1}$, the elements of $A$ and $x$ are differentiable functions of $\theta$.

What is the derivative of the product $Ax$ with respect to the parameter $\theta$ :

$$\frac{\partial Ax}{\partial \theta}?$$

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Let's denote the Matrix $A = \begin{pmatrix} g_{1,1}(\theta) & g_{1,2}(\theta), ...\\ g_{2,1}(\theta) & g_{2,2}(\theta), ...\\.\\.\\.\end{pmatrix}$.

Then it holds $Ax = \begin{pmatrix}g_{1,1}(\theta)\cdot x_1(\theta) + g_{1,2}(\theta)\cdot x_2(\theta) + ...\\ g_{2,1}(\theta)\cdot x_1(\theta) + g_{2,2}(\theta)\cdot x_2(\theta) + ...\\.\\.\\.\end{pmatrix}$.

You can easily differentiate every summand of every entry of the vector, w.r.t to $\theta$ by product rule, to get (I dropped "$(\theta)$" here for simple notation) $$ \frac{\partial Ax}{\partial \theta} = \begin{pmatrix} (g'_{1,1} \cdot x_1+ g_{1,1} \cdot x_1')+ (g'_{1,2}\cdot x_2 + g_{1,2} \cdot x_2')+ ...\\ (g'_{2,1} \cdot x_1+ g_{2,1} \cdot x_1')+ (g'_{2,2}\cdot x_2 + g_{2,2} \cdot x_2')+ ...\\ .\\.\\. \end{pmatrix} $$

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    Laray, the entries of $x$ are also depend on $\theta$, I guess you did not notice that. – 2017-01-18
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    Yep, was an oversight on that, be right back with results... – 2017-01-18
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    user137684 And here they are! – 2017-01-18
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    So from your response, we can write $\frac {\partial Ax}{\partial \theta} = \frac {\partial A}{\partial \theta}x+A \frac {\partial x}{\partial \theta}$. – 2017-01-18
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    I did not realise this, but yes, you are right. Seems like product rule works in matric-calculus as well... – 2017-01-18