I have seen the previous post and I tried to follow the answer to solve the problem. But I encountered some problem and I can not get the expected answer $2a_{ij}$. I stuck at the final line since I thought when $k=j$ , we get $da_{ij}a_{ji}+a_{ij}(da_{jj})$.
$$(AA^T)ij=∑_{k=1}^{n}a_{ik}a_{jk}$$ \begin{align} d((AA^T)ij)&=d(∑_{k=1}^{n}a_{ik}a_{jk}) \\ &=∑_{k=1}^{n}[(da_{ik})a_{jk}+a_{ik}(da_{jk})] \\ \end{align}