TITLE: Derivative of $\log |AA^T|$ with respect to $A$
QUESTION [5 upvotes]: What is the derivative of $\log |AA^T|$ with respect to $A$, where $|A|$ denotes the determinant of $A$?

REPLY [2 votes]: Let $M = AA^T$ and $f = \log \det \left( M \right)$.
We will utilize the following the identities 

Trace and Frobenius product relation $$A:B={\rm tr}(A^TB)$$ or $$A^T:B={\rm tr}(AB)$$
Cyclic property of Trace/Frobenius product $$\eqalign{
A:BC
 &= AC^T:B \cr
 &= B^TA:C \cr
 &= {\text etc.} \cr
}$$
Jacobi's formula (for nonsingular matrix $M$)  $$\log \det \left( M \right) = {\rm tr}\log\left( M \right)$$

Now, we obtain the differential first and thereafter we obtain the gradient. 
So, 
\begin{align}
df &= d \log \det \left( M \right) \\
   &= d \ {\rm tr}\left( \log\left( M \right) \right) \hspace{8mm} \text{note: utilized Jacobi's formula} \\   
   &= {\rm tr} \left( M^{-1} dM \right)  \\
   &= M^{-T} \ : \ dM \hspace{8mm} \text{note: utilized trace and Frobenius relation} \\
   &= \left(AA^T\right)^{-T} \ : \ d\left(AA^T\right)  \\   
  &= \left(AA^T\right)^{-1} \ : \ d\left(AA^T\right)  \\ 
  &= \left(AA^T\right)^{-1} \ : \ \left( dA \ A^T + A \ dA^T\right)   \\ 
  &= \left(AA^T\right)^{-1}  : 2 \ dA \ A^T   \\
  &=  2 \left(AA^T\right)^{-1} (A^T)^T  : dA   \hspace{8mm} \text{note: utilized cyclic property of Frobenius product} \\
  &=  2 \left(AA^T\right)^{-1} A  : dA   \\
\end{align}
So, the derivative of $f = \log \det \left( AA^T \right)$ with respect to $A$ is 
\begin{align}
\frac{\partial}{\partial A} f 
= \frac{\partial}{\partial A} \log \det \left(  AA^T \right) 
=  2 \left(AA^T\right)^{-1} A  .\\
\end{align}