Derivative of trace involving the Hadamard product I am trying to find a matrix product of a co-occurrence matrix
$$\mathbf{Y} \approx \mathbf{U} \mathbf{V}^\intercal,$$ where $\mathbf{Y} \in \mathbb{R}^{I \times J}$, $\mathbf{U} \in \mathbb{R}^{I \times K}$ and $\mathbf{X} \in \mathbb{V}^{J \times K}$.
To find this, I am going to minimise the error (note I have added regularisation terms to prevent overfitting)
$$\xi(\mathbf{U},\mathbf{V}) =  ||\mathbf{Y} - \mathbf{U} \mathbf{V}^\intercal ||^2_F + \lambda_u || \mathbf{U} \mathbf{U}^\intercal ||^2_F + \lambda_v || \mathbf{V} \mathbf{V}^\intercal ||^2_F. $$
Then to minimise this I will find the derivatives with respect to both $\mathbf{U}$ and $ \mathbf{V} $:
$$\nabla_\mathbf{U} = -2\mathbf{V} \left( \mathbf{Y} - \mathbf{U}\mathbf{V}^{ \intercal} \right) + 2\lambda_u \mathbf{U}$$
$$\nabla_\mathbf{V} = -2\mathbf{U}\left( \mathbf{Y} - \mathbf{U}\mathbf{V}^\intercal \right) + 2\lambda_v \mathbf{V}.$$
These have been calculated using the identities:
$$ ||\mathbf{A}||^2_F = \text{tr}(\mathbf{A}\mathbf{A}^\intercal) $$ and $$ \frac{\partial}{\partial \mathbf{Y}} \text{tr}((\mathbf{A}\mathbf{Y} + \mathbf{C})(\mathbf{A}\mathbf{Y} + \mathbf{C})^\intercal) = \mathbf{A^\intercal(\mathbf{A}\mathbf{Y} + \mathbf{C})}.$$
I have no issue with these derivations, the problem is I am now trying to solve the same problem but with an added $\textit{masking matrix}$ (for reasons I won't go into here), which is a randomly populated binary matrix $\mathbf{M} \in \mathbb{R}^{I \times J}$. This changes the error function to
$$\xi(\mathbf{U},\mathbf{V}) =  ||\mathbf{M}\odot(\mathbf{Y} - \mathbf{U} \mathbf{V}^\intercal) ||^2_F + \lambda_u || \mathbf{U} \mathbf{U}^\intercal ||^2_F + \lambda_v || \mathbf{V} \mathbf{V}^\intercal ||^2_F. $$
I'm unsure how I would find the derivative of this new error function as the Hadamard product complicates things - if anyone could shed some light on this it would be much appreciated!
Thank you.

$\def\p{\partial}$
The binary matrix $M$ has the property that $\;M\odot M = M$
Let a colon denote the trace/Frobenius product, i.e.
$$B:C={\rm Tr}(B^TC)={\rm Tr}(C^TB)=C:B$$
The Hadamard and Frobenius products commute in the following senses
$$\eqalign{
A:B &= B:A \\
A\odot B &= B\odot A \\
C:(A\odot B) &= (C\odot A):B \\
}$$
For typing convenience define the matrices
$$\eqalign{
X &\doteq UV^T-Y \\
W &\doteq M\odot X \\
}$$
Notice that
$$\eqalign{
M\odot W &= M\odot M\odot X \\&= M\odot X \\&= W
}$$
Write the error function in terms of these and calculate its differential.
$$\eqalign{
{\cal E} &= W:W + \lambda_uU:U + \lambda_vV:V \\
\\
d{\cal E}
 &= 2W:dW + 2\lambda_uU:dU + 2\lambda_vV:dV \\
 &= 2W:(M\odot dX) + 2\lambda_uU:dU + 2\lambda_vV:dV \\
 &= 2W:(dU\,V^T+U\,dV^T) + 2\lambda_uU:dU + 2\lambda_vV:dV \\
 &= 2WV:dU + 2U^TW:dV^T + 2\lambda_uU:dU + 2\lambda_vV:dV \\
 &= 2(WV+\lambda_uU):dU
  \;+\; 2(W^TU+\lambda_vV):dV \\
}$$
From which the gradients can be identified as
$$\eqalign{
\frac{\p{\cal E}}{\p U} &= 2(M\odot X)V + 2\lambda_uU \\
\frac{\p{\cal E}}{\p V} &= 2(M\odot X)^TU + 2\lambda_vV \\
}$$
Note that setting $M=J\,$ (i.e. the all-ones matrix) recovers the gradients in the original (non-Hadamard) case, since for any matrix $A$ one has $\;J\odot A = A$.
