I saw in this post that
$\frac{d}{dt}\text{logm}(Z(t)) = \frac{dZ(t)}{dt}(Z(t))^{-1}$
Is this true to say:
$\frac{d}{{dU}}{\mathop{\rm logm}\nolimits} (A) = {A^{ - 1}}\frac{d}{{dU}}A$
where U is an m by n matrix and A is an m by m matrix which is a function of U ?
EDIT: A is not Symmetric and positive definite.
Neither of those statements is true.
However, you can use a block triangular matrix to calculate the Frechet derivative using the method of Kenney & Laub $${\rm F}\Bigg(\begin{bmatrix}Z&E\\0&Z\end{bmatrix}\Bigg) = \begin{bmatrix}F&L\\0&F\end{bmatrix} $$ where $F = {\rm F}(Z)$ and $$L = \lim_{dt\rightarrow\,0}\, \frac{{\rm F}(Z+E\,dt)-{\rm F}(Z)}{dt}$$ For your particular case
$$\eqalign{ F &= {\rm logm}(Z) \cr E &= \frac{dZ}{dt} \cr }$$ $Z$, $F$ and $E$ must be evaluated at the same moment in time, e.g. $t=t_0$.
This will give you the directional derivative of the function, in the $E$-direction.