Derivative of quadratic form of matrix in terms of the matrix elements?

Question

Suppose I have $b^tAc$ and I try to get the derivative in terms of $A$. How could What is the matrix notational result?

I believe the answer is $bc^t$, isn't it ?

Answer 1

Effectively, the answer is $bc^t$.

The derivative of a scalar with respect to a matrix is calculated as

$\frac{\partial y}{\partial \mathbf{A}} = \begin{bmatrix} \frac{\partial y}{\partial a_{11}} & \frac{\partial y}{\partial a_{21}} & \cdots & \frac{\partial y}{\partial a_{p1}}\\ \frac{\partial y}{\partial a_{12}} & \frac{\partial y}{\partial a_{22}} & \cdots & \frac{\partial y}{\partial a_{p2}}\\ \vdots & \vdots & \ddots & \vdots\\ \frac{\partial y}{\partial a_{1q}} & \frac{\partial y}{\partial a_{2q}} & \cdots & \frac{\partial y}{\partial a_{pq}}\\ \end{bmatrix}$,

where one have to take into account that the indices of $A$ elements are transposed in the resulting matrix.