Prove matrix derivative identity I found the following identity in the Matrix Cookbook:
\begin{align*}
\frac{\partial }{\partial X} \text{Trace}\left( X B X ^{\mathrm{T}} \right) = XB^{\mathrm{T}} + XB,
\end{align*}
where $X\in\mathbb{R}^{n\times m}$ and $B\in\mathbb{R}^{m \times m}$. Any hints on how I prove this?

A solution avoiding to use the coordinates.
Note $f : \mathbb{R}^{n\times m} \to \mathbb R$ where
$$f(X) = \text{Trace}\left( X B X ^{\mathrm{T}} \right)$$
You have $f=h \circ g$ where $g(X) = X B X ^{\mathrm{T}}$ and $h(Y) = \text{Trace}(Y)$.
According to the chain rule you have
$$(h \circ g)^\prime (X) = (h^\prime(g(X)) \circ g^\prime(X)$$
$h$ is a linear map. So its (Fréchet) dérivative is
$$h^\prime(Y).V = \text{Trace}(V)$$
On its side, $g$ is a bilinear map. So we have
$$g^\prime(X).U = U B X ^{\mathrm{T}} + X B U ^{\mathrm{T}}$$
Using the chain rule with the above derivatives, you get
$$f^\prime(X).U = \text{Trace}(U B X ^{\mathrm{T}} + X B U ^{\mathrm{T}})$$
As $\text{Trace}$ is linear and $\text{Trace}(X)=\text{Trace}(X^{\mathrm{T}})$ for all square matrix, we also have
$$f^\prime(X).U = \text{Trace}((X B^{\mathrm{T}}  + X B)U ^{\mathrm{T}})$$
Which is exactly the result you're looking for.
