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What is the result of

$$ \frac{\rm d}{{\rm d}t}\Big(A\left(x(t)\right)\cdot y\Big) $$

where variables $t\in\mathbb R$ and $y\in\mathbb R^n$, functions $x\colon=\mathbb R\to\mathbb R^\ell$ and $A\colon=\mathbb R^\ell\to\mathbb R^{m\times n}$.

Thanks in advance!

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    What are your thoughts? If you show some effort people are more likely to help. – 2017-02-28
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    @Rumplestillskin Sorry I am new here. Now I try my own answer below, can you give any suggestions? And can you click the up button to give encouragement. Thanks. – 2017-02-28
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    Is $y$ a function of $t$? – 2017-02-28
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    @Sid Thanks for your interest. No, $y$ is not a function of $t$, it is a variable. – 2017-02-28

1 Answers 1

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$$ \left(y\otimes I_{m\times m}\right)^{\mathsf T}\cdot D\cdot\dot x $$ where $D\in\mathbb R^{mn\times\ell}$ and the $i$-th column of $D$ is the derivative of ${\rm vec}(A)$ with respect to $x_i$ for $i=1,2,\dots,\ell$.

This answer is motivated by the accepted answer of the question What is the result of this type of derivative?

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    Should it not be $y \otimes I$ instead of $I \otimes y$? The rule is ${\rm vec}(B X C) = (C^{\rm T} \otimes B) \cdot {\rm vec}(X)$. In your case, $B=I$, $X=A$, and $C=y$. Otherwise looks good. Of course, you can pack the vectorized partial derivatives as columns of a matrix (say, $D$) and the ${\rm d} x_i/{\rm d}t$ into a vector (say, $\dot{x}$) to simplify things further (namely, into $(y\otimes I)^{\rm T} \cdot D \cdot \dot{x}$). – 2017-02-28
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    @Florian I modify my answer according to your suggestion. It does help. Thank you so much again! – 2017-03-01
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    @Florian A further question: the result of the original problem $\frac{\rm d}{{\rm d}t}\left(A\left(x(t)\right)\cdot y\right)\in\mathbb R^m$, but why my answer (and is also yours) is a scalar, $\left(y\otimes I_{m\times m}\right)^{\mathsf T}\cdot D\cdot\dot x\in\mathbb R$? – 2017-03-01
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    No it's not scalar: $y$ is $n \times 1$ so $y \otimes I_m$ is $nm \times m$. Moreover, $D$ is $mn \times \ell$ and $\dot{x}$ is $\ell \times 1$, so the whole product is $(m \times mn) \cdot (mn \times \ell) \cdot(\ell \times 1) = (m \times 1)$, as expected. – 2017-03-01
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    @Florian Yes, you are right, sorry for my carelessness. Thank you! – 2017-03-01