a practical question about matrix derivative with inverse and chain rule: dimension mismatch Recently, I was trying to take the following derivative 
$$
\dfrac{\partial (X^TV^{-1}X)^{-1}}{\partial V} 
$$
I was referring to matrix cookbook to solve it, where I found several useful equations:    
Equation (59) says:
$$
\dfrac{\partial Y^{-1}}{\partial x} = -Y^{-1}\dfrac{\partial Y}{\partial x}Y^{-1}
$$
so, I think I have:
$$
\dfrac{\partial (X^TV^{-1}X)^{-1}}{\partial V^{-1}} = -(X^TV^{-1}X)^{-1} X^TX(X^TV^{-1}X)^{-1}
$$
and 
$$
\dfrac{\partial V^{-1}}{\partial V} = -V^{-1}V^{-1}
$$
According to the chain rule, it should be:
$$
\dfrac{\partial (X^TV^{-1}X)^{-1}}{\partial V} =\dfrac{\partial (X^TV^{-1}X)^{-1}}{\partial V^{-1}}\dfrac{\partial V^{-1}}{\partial V} = ((X^TV^{-1}X)^{-1} X^TX(X^TV^{-1}X)^{-1})^T V^{-1}V^{-1}
$$
However, I met one problem. $V$ is a matrix of size $(n, n)$ and $X$ is a matrix of size $(n, m)$. 
Then, the first half of the chain rule is of size of $(m, m)$, while the second half of the chain rule is of size $(n, n)$. 
Please help me figure out what goes wrong. 
Thanks ahead. 
 A: Let's define some intermediate variables
$$\eqalign{
 P &= V^{-1} \cr
 M &= X^TPX \cr
 F &= M^{-1} \cr
}$$
whose differentials are
$$\eqalign{
 dP &= -V^{-1}\,dV\,V^{-1} \cr
 dM &= X^T\,dP\,X \cr
 dF &= -M^{-1}\,dM\,M^{-1} \cr
}$$
That last differential is the one we're interested in, so let's successively substitute variables until we get back to $V$
$$\eqalign{
 dF &= -M^{-1}\,dM\,M^{-1} \cr
   &= -M^{-1}\,(X^T\,dP\,X)\,M^{-1} \cr
   &= -M^{-1}\,X^T\,(-V^{-1}\,dV\,V^{-1})\,X\,M^{-1} \cr
   &= M^{-1}\,X^T\,V^{-1}\,dV\,V^{-1}\,X\,M^{-1} \cr
   &= F\,X^T\,V^{-1}\,dV\,V^{-1}\,X\,F \cr
}$$
At this point, let's follow the prescription of Magnus & Neudecker for dealing with matrix-by-matrix derivatives, and vectorize both sides
$$\eqalign{
 d{\rm vec}(F) &= (V^{-1}\,X\,F)^T\otimes(F\,X^T\,V^{-1})\,d{\rm vec}(V) \cr
}$$
Which can be rearranged to the conventional looking result
$$\eqalign{
 \frac{\partial f}{\partial v} &= (V^{-1}\,X\,F)^T\otimes(F\,X^T\,V^{-1}) \cr
}$$
A: I don´t see how equation (59) should help. $V$ does not depend on $X$.
If $X$ is a square matrix then $\left(X^TV^{-1}X \right)^{-1}=X^{-1}V\left(X^{-1}\right)^T$
Let $X^{-1}=A$. We get $AVA^T$
You can see here (page 9) that $$\frac{\partial AVA^T}{\partial V}= AA^T$$
