Consider $J = \|X - AS\|^2$, where $A \in \mathbb{R}^{N \times R}$, $S \in \mathbb{R}^{R \times M}$.
We need to calculate $\dfrac{d J}{d S}$.
I've done the following:
$J = \operatorname{tr}((X - AS)(X - AS)^\top) = \operatorname{tr}((X-AS) (X^\top - S^\top A^\top)) = \operatorname{tr}(XX^\top - ASX^\top - XS^\top A^\top + ASS^\top A^\top)$ and hence
$$\dfrac{dJ}{dS} = \dfrac{-\operatorname{tr}(ASX) - \operatorname{tr}(XS^\top A^\top) + \operatorname{tr}(ASS^\top A^\top)}{dS}$$
$\dfrac{d\operatorname{tr}(ASX)}{dS} = \dfrac{d\operatorname{tr}(SXA)}{dS} = (XA)^\top = A^\top X^\top$
$\dfrac{d\operatorname{tr}(XS^\top A^\top)}{dS} = \dfrac{d\operatorname{tr}((XS^\top A^\top)^\top}{dS} = \dfrac{d\operatorname{tr}(ASX^\top)}{dS} = \dfrac{\operatorname{tr}(SX^\top A)}{dS} = (X^\top A)^\top = A^\top X$
$\dfrac{d\operatorname{tr}(AS(AS)^\top)}{dS}$, so $$\displaystyle \dfrac{d\sum_{n,m}(AS)_{n,m} ((AS)^\top)_{m,n}}{d s_{i,j}} = \dfrac{d\sum_{n,m}(AS)^2_{n,m}}{d s_{i,j}} = \dfrac{d \sum_{n,m} \left(\sum_{q} a_{n,q} s_{q,m}\right)^2}{d s_{i,j}} = \dfrac{d\sum_{n,m} \sum_{q}a_{n,q}s_{q,m} \sum_{l}a_{n,l} s_{l,m}}{ds_{i,j}} =$$ $$\sum_{n,m} \sum_{q} \sum_{l} a_{n,q} s_{q,m} a_{n,l} \delta_{l}^{i} \delta_{m}^{j} = \sum_{n} \sum_{q} a_{n,q}s_{q,j} a_{n, i} = \sum_{n} A^\top_{i,n}(AS)_{n,j} = (A^\top A S)_{i,j}$$
Hence we have: $\dfrac{dJ}{dS} = A^\top (AS - X - X^\top)$. Unfortunately, I'm very bad at matrix calculus and do not use some kind of properties. I'm interested in whether my approach is OK?
We have, using $\operatorname{tr}(C+C^T)=2\operatorname{tr}(C)$ $$\begin{align} d_SJ(v)&=\operatorname{tr}\bigl((-Av)^T(X-AS)+(X-AS)^T(-Av)\bigr)\\ &=\operatorname{tr}\bigl((X-AS)^T(-Av)+((X-AS)^T(-Av))^T \bigr)\\ &=-2\operatorname{tr}\bigl((X-AS)^TAv\bigr)\\ &=-2\operatorname{tr}\bigl(\bigl(A^T(X-AS)\bigr)^Tv\bigr)\\ &=-2\langle (A^T(X-AS)),v\rangle. \end{align} $$ Hence $$d_SJ=-2(A^T(X-AS)).$$
EDIT (even simpler) Using chain and product rule: $$d_SJ(v)=2\langle X-SA,-vA\rangle=-2\langle (A^T(X-AS)),v\rangle.$$