Derivative of $\log |AA^T|$ with respect to $A$ What is the derivative of $\log |AA^T|$ with respect to $A$, where $|A|$ denotes the determinant of $A$?
 A: Let $f : \mathbb{R}^{m \times n} \to \mathbb{R}$ be defined by
$$f (X) = \log |X X^T|$$
The directional derivative of $f$ in the direction of $V \in \mathbb{R}^{m \times n}$ is
$$\begin{array}{rl} D_V f (X) &= \displaystyle\lim_{h \to 0} \frac{1}{h} \left( \log |(X + h V) (X + h V)^T| - \log |X X^T| \right)\\\\ &= \displaystyle\lim_{h \to 0} \frac{1}{h} \left( \log |X X^T + h V X^T + h X V^T + h^2 V V^T| - \log |X X^T| \right)\end{array}$$
If $X$ has full row rank, i.e., if $\operatorname{rank} (X) = m$, then $X X^T$ is invertible. Hence,
$$\begin{array}{rl} D_V f (X) &= \displaystyle\lim_{h \to 0} \frac{1}{h} \left( \log |X X^T + h V X^T + h X V^T + h^2 V V^T| - \log |X X^T| \right)\\\\ &= \displaystyle\lim_{h \to 0} \frac{1}{h} \left( \log |X X^T| + \log | I_m + h (X X^T)^{-1} \left( V X^T + X V^T + h V V^T \right)| - \log |X X^T| \right)\\\\ &= \displaystyle\lim_{h \to 0} \frac{1}{h} \, \log | I_m + h (X X^T)^{-1} \left( V X^T + X V^T + h V V^T \right)|\\\\ &= \displaystyle\lim_{h \to 0} \frac{1}{h} \, \log ( 1 + h \operatorname{tr} ((X X^T)^{-1} \left( V X^T + X V^T + h V V^T \right)))\\\\ &= \displaystyle\lim_{h \to 0} \operatorname{tr} ((X X^T)^{-1} \left( V X^T + X V^T + h V V^T \right))\\\\ &= \operatorname{tr} ((X X^T)^{-1} \left( V X^T + X V^T \right))\\\\ &= 2 \, \operatorname{tr} (X^T (X X^T)^{-1} V)\end{array}$$

matrix-calculus scalar-fields determinant logarithms
A: Assume that $A\in M_{n,m}$ where $n\leq m$ and $rank(A)=n$ and let $f(A)=\log(\det(AA^T))$. Then $Df_A:H\rightarrow tr((AA^T)^{-1}(HA^T+AH^T))$ or, if $A$ is a derivable matrix function, then $f'(A)=tr((AA^T)^{-1}(A'A^T+AA'^T))$.
A: 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}
A: Hint 1:
$\partial \log\det(M)=trace(M^{-1} \partial M)$
Hint 2:
Use the chain rule.
I hope this helps. Please let us know if you couldn't figure it out.
