How to get the gradient of a matrix product with respect to a vector? I am new to vector calculus and even though I was able to understand how to derive simpler functions like the product of two vectors in $\mathbb{R}^m$, I am confused about this specific scenario.
Lets say we have a column vector $x \in \mathbb{R}^m, A \in \mathbb{R}^{m\times m}$.
How can I determine the gradient of $x^TA$ with respect to $x$?
I am not able to grasp how you derive something with respect to a vector.
From what I can see $x^TA$ is equal to $[A_1 \cdot x$, $A_2 \cdot x$, ..., $A_m \cdot x]$ for all columns of $A$. Which means the operation yields a row vector in $\mathbb{R}^m$. 
I tried to derive this solution with respect to the individual scalars of $x$ but I am lost since I am clearly not approaching it correctly.
How would one solve this?

If you now look at $x^TA$ as a linear map $\mathbb{R}^m\to \mathbb{R}^m$ as $$f(x_1,\ldots, x_m)=(\sum_{i=1}^m a_{i1}x_i,\ldots,\sum_{i=1}^m a_{im}x_i)$$ so we don't have to think about $f$ in columns and rows or matrices, we know that the gradient of the linear map is written as
$$\frac{\partial f}{\partial x}=\left(\begin{array} 
\ a_{11} & a_{21} & \dots  & a_{m1} \\
    a_{12} & a_{22} & \dots  & a_{m2} \\
    \vdots & \vdots & \ddots & \vdots \\
    a_{1m} & a_{2m} & \dots  & a_{mm}
\end{array}\right)=A^T$$
by convention of derivatives are from left to right and $f$-elements from the top down.
