I have a doubt, I am not able to come up with a solution.
I have 3 vectors: $\vec{a},\vec{b},\vec{c} \in \mathbb{R}^{3\times1}$
I want to compute the derivative:
$H=\frac{\partial \vec{a}^T \vec{b}}{\partial \vec{c}}$ which should be $\in \mathbb{R}^{1\times3}$ because $\vec{a}^T \vec{b}\in \mathbb{R}^{1\times1}$
If I apply the chain rule to $H$ I get:
$H=(\frac{\partial\vec{a}}{\partial\vec{c}})^T\vec{b}+\vec{a}^T(\frac{\partial\vec{b}}{\partial\vec{c}})$
An hypothesis of my problem is that $(\frac{\partial\vec{b}}{\partial\vec{c}})=0$, so $H=(\frac{\partial\vec{a}}{\partial\vec{c}})^T\vec{b}$ which is $\in\mathbb{R}^{3\times1}$.
How can I get something which is $\in\mathbb{R}^{1\times3}$? The first thing that came to my mind is just to put a transpose in the chain rule formula like that:
$H=[(\frac{\partial\vec{a}}{\partial\vec{c}})^T\vec{b}]^T+\vec{a}^T(\frac{\partial\vec{b}}{\partial\vec{c}})$
But I am not able to justify this operation.
Thanks for your help.
Let $$\eqalign{ A &= \frac{\partial a}{\partial c}, &\, B &= \frac{\partial b}{\partial c} \cr }$$ Then $$\eqalign{ \frac{\partial\,(a^Tb)}{\partial c} &= A^Tb + B^Ta \cr }$$