How to prove $\frac{\partial}{\partial{W}} \operatorname{trace}((Y-XW)(Y-XW)^T)=2X^T(XW-Y)$? $\newcommand{\tr}{\operatorname{tr}}$ I want to prove the following expression  with simple matrix operations.
\begin{align}
& \frac{\partial}{\partial{W}} \tr((Y-XW)(Y-XW)^T)=2X^T(XW-Y) \\[10pt]
= {} & \frac{\partial}{\partial{W}} \tr((Y-XW)(Y-XW)^T) \\[10pt]
= {} & \frac{\partial}{\partial{W}} \tr(YY^T - YW^TX^T - XWY^T + XW(XW)^T) \\[10pt]
= {} & \frac{\partial}{\partial{W}} \tr( -YW^TX^T - XWY^T + XW(XW)^T) \\[10pt]
= {} & -2X^TY  +\frac{\partial}{\partial{W}} \tr(XW(XW)^T)
\end{align}
Now I need to calculate  $ \dfrac{\partial}{\partial{W}} \tr(XW(XW)^T)$.
Can anyone help to calculate this derivate?
Thanks.
=========
based on the suggestion to write the expression in Einstein notation I found:
$\begin{align}
&\dfrac{\partial}{\partial{W}} \tr(XW(XW)^T)=\\[10pt]
= {} &   \dfrac{\partial}{\partial{W}} \tr(XW(XW)^T)\\[10pt]
= {} &  \dfrac{\partial}{\partial{W}} \sum_i\sum_j\sum_k\sum_l X_{ij}W_{jk}W^T_{kl}X^T_{li}\\[10pt]
= {} &  \dfrac{\partial}{\partial{W_{jk}}} \sum_i\sum_j\sum_k\sum_l X_{ij}W_{jk}W^T_{kl}X^T_{li} + \dfrac{\partial}{\partial{W_{lk}}} \sum_i\sum_j\sum_k\sum_l X_{ij}W_{jk}W_{lk}X^T_{li} \\[10pt]
= {} & \sum_i\sum_k\sum_l(W^TX^T)_{ki} X_{ij}   +  \sum_i\sum_j\sum_k X_{ij}W_{jk} X^T_{li}  \\[10pt]
= {} & \sum_i\sum_k\sum_l(W^TX^T)_{ki} X_{ij}   +  \sum_i\sum_j\sum_kX^T_{li} X_{ij}W_{jk} \\[10pt]
={} &2X^TXW  
\end{align}$
Please let me know if something is wrong with this. Thanks.
 A: write everything in einstein notation, and the result follows steadily.
Or, exploit the great flexibility of the concept of differential:
$$
d_{W}\operatorname{tr}((Y-XW)(Y-XW)^T)=\operatorname{tr}(d_{W}[(Y-XW)(Y-XW)^T])=
$$
$$
=\operatorname{tr}\left([d_{W}(Y-XW)][(Y-XW)^T]+[(Y-XW)][d_{W}(Y-XW)^T]\right)=
$$
$$
=\operatorname{tr}\left((-X\,d_{W}W)(Y-XW)^T+(Y-XW)(-X\,d_{W}W)^T\right)=
$$
$$
=\operatorname{tr}\left((-X\,d_{W}W)(Y-XW)^T+(-X\,d_{W}W)(Y-XW)^T\right)=
$$
$$
=2\operatorname{tr}\left((-X\,d_{W}W)(Y-XW)^T\right)=
$$
$$
=2\operatorname{tr}\left((XW-Y)^T(X\,d_{W}W)\right)=
$$
$$
=2\operatorname{tr}\left(d_{W}W^T\, X^T(XW-Y)\right)
$$
so:
$$
d_{W}\operatorname{tr}((Y-XW)(Y-XW)^T)=2\operatorname{tr}\left(d_{W}W^T\, X^T(XW-Y)\right)=\#
$$
using index notation:
$$
\#=2\sum_i\left[d_{W}W^T\, X^T(XW-Y)\right]_{i,i}=
$$
$$
=\sum_i\sum_q\left[d_{W}W^T\right]_{i,q}\left[2X^T(XW-Y)\right]_{q,i}=
$$
$$
=\sum_i\sum_q\left[d_{W}W\right]_{q,i}\left[2X^T(XW-Y)\right]_{q,i}
$$
so:
$$
\frac{d}{dW_{q,i}}\operatorname{tr}((Y-XW)(Y-XW)^T)=\left[2X^T(XW-Y)\right]_{q,i}
$$
A: I didn't learn about matrix calculus, so I use the notation of matrix calculus in Wikipedia
$$d(tr(XW(XW)^T))=tr(d(XWW^TX^T))$$
$$=tr(d(XW)W^TX^T+XWd(W^TX^T))$$
$$=tr(X(dW)W^TX^T)+tr(XW(dW^T)X^T)$$
$$=tr(X(dW)W^TX^T)+tr((dW)^TX^TXW)$$
$$=tr(W^TX^TX(dW))+tr((X^TXW)^TdW)$$
$$=tr(2(X^TXW)^TdW)$$
$$\therefore \frac{\partial}{\partial W} tr(XW(XW)^T)=2X^TXW$$
A: The Frobenius product is a convenient way to denote the trace  $\,\,A:BC={\rm tr}(A^TBC)$
Rules for rearranging terms in a Frobenius product follow directly from properties of the trace.
Let $Z=(XW-Y)$, then find the differential and gradient of the function as 
$$\eqalign{
 f &= Z:Z \cr
df &= 2Z:dZ = 2Z:X\,dW = 2X^TZ:dW \cr
\frac{\partial f}{\partial W} &= 2X^TZ = 2X^T(XW-Y) \cr
}$$
