Differentiating a vector multiplied with a matrix I was reading about differentiating a column vector $x$ multiplied with matrix $A$. I have found this formula:
$\frac{\partial x^TA}{\partial x} = \frac{\partial A^Tx}{\partial x} = A$
Is this correct?
How can $x$ just disappear from the differentiation?
 A: It is correct, because $$\frac{\partial x^Ta}{\partial x}$$ is defined as the vector where the $i$-th quantity is equal to $$\frac{\partial x^Ta}{\partial x_i}$$
and since $$x^Ta = x_1a_1 + x_2a_2+\cdots +x_n a_n$$
you cans see that $$\frac{\partial x^Ta}{\partial x_i} = a_i.$$

This means that the $i$-th component of $$\frac{\partial x^Ta}{\partial x}$$ is equal to $a_i$, which means $$\frac{\partial x^Ta}{\partial x}$$ is equal to $$a.$$
A: Hint:
It is a special case of the derivative of a vector function $\vec y=\vec y(\vec x)$. For such kind of functions the derivative can be defined as the ''linear function'' that locally approximate the given function. (This is the Frechet derivative and you can see the exact definition here).
In the case $\vec y=\vec y(\vec x)$ the derivative is defined as the matrix (Jacobian matrix):
$$
\frac{\partial \vec y}{\partial \vec x}=\begin{bmatrix}
\frac{\partial y_1}{\partial  x_1}&\cdots &\frac{\partial y_1}{\partial  x_n}\\
\frac{\partial y_2}{\partial  x_1}&\cdots &\frac{\partial y_2}{\partial  x_n}\\
\cdots\\
\frac{\partial y_m}{\partial  x_1}&\cdots &\frac{\partial y_m}{\partial  x_n}\\
\end{bmatrix}
$$
You can easily see that if $\vec y=A \vec x$ that Jacobian matrix is just $A$
