Is there a simple identity for the derivative of a matrix logarithm w.r.t. a real parameter? Let $A(t)$ be an invertible square matrix that depends (differentiably) on a real parameter $t$.
It is well known that for example
$$
\frac{d}{dt} A(t)^{-1}=-A(t)^{-1}\ \dot{A}(t)\ A(t)^{-1}
$$
and
$$
\frac{d}{dt} \text{det}\ A(t)=\text{det}( A(t))\ \text{tr}(A(t)^{-1}\dot{A}(t)).
$$
Is there a similarly simple identity for
$$
\frac{d}{dt}\log A(t)=?
$$
 A: First, note that the chain rule for matrix functions (i.e. functions which map matrices to matrices) results in a rank-4 tensor:
$$
\frac{d}{dt}F(A(t))_{ab} = \sum_{cd} F'(A(t))_{ab;cd} \frac{dA(t)_{cd}}{dt}
$$
where $F'(A(t))$ is a rank-4 tensor which encodes the derivative of $F$ and $a$, $b$, $c$, and $d$ are indices of the above matrices and tensors. For example, if $F(A) = A^{-1}$, then
$$
F'(A(t))_{ab;cd} = - (A(t)^{-1})_{ac} (A(t)^{-1})_{db}
$$
which reproduces the expression for $\frac{d}{dt}A(t)^{-1}$ given in the question.
For the case $F = \log$ and if $A(t)$ is diagonalizable with no eigenvalues that are zero or on the negative real axis (i.e. the principal branch cut of $\log$), then the answer is given on page 146 (see 2nd to last equation) of Jog, C.S. J Elasticity (2008) 93: 141. doi:10.1007/s10659-008-9169-x and can be expressed as
$$
\log'(A(t))_{ab;cd} = \sum_{ij} P^{(i)}_{ac} P^{(j)}_{db}
  \begin{cases} 
    \lambda_i^{-1} & \lambda_i = \lambda_j \\
    \frac{\log\lambda_i - \log\lambda_j}{\lambda_i - \lambda_j} & \lambda_i \neq \lambda_j
  \end{cases}
$$
where $i$ and $j$ index the eigenvalues $\lambda$ of $A(t)$, and $P^{(i)}_{ab} \equiv Q_{ai} (Q^{-1})_{ib}$ projects onto the $i$-th eigenvector where $Q$ is the matrix of eigenvectors of $A(t)$ given by the eigendecomposition $A(t) = Q \Lambda Q^{-1}$. Therefore
$$
\frac{d}{dt}\log A(t) = \sum_{ij} P^{(i)} \cdot \frac{dA(t)}{dt} \cdot P^{(j)}
  \begin{cases} 
    \lambda_i^{-1} & \lambda_i = \lambda_j \\
    \frac{\log\lambda_i - \log\lambda_j}{\lambda_i - \lambda_j} & \lambda_i \neq \lambda_j
  \end{cases}
$$
(I checked this equation in a Mathematica notebook.)
A: A common definition of the logarithm for (finite dimensional) matrices is via the Dunford-Taylor integral:
$$
\ln(T) := \frac{1}{2\pi i} \oint_\Gamma \ln(z) (z-T)^{-1} dz \, ,
$$
Where $\Gamma$ is a simple closed smooth curve that contains all the eigenvalues of $T$ in anticlockwise direction. Note that the above is a seemingly natural generalization of Cauchy's integral formula. 
Let the matrix $T$ depend on some parameter $x$. As you noted yourself one has
$$
\frac{\partial }{\partial x} (z - T)^{-1}= (z - T(x))^{-1} T' (z- T(x))^{-1} \, , 
$$ 
where prime denotes differentiation with respect to $x$.  
Combining the two equations we get 
$$
\frac{\partial \ln(T)}{\partial x} = \frac{1}{2\pi i} \oint_\Gamma \ln(z) (z - T(x))^{-1} T' (z- T(x))^{-1} dz \, .
$$
a good reference is the book of Kato Perturbation theory for linear operators. 
A: First, let us do not care about convergence issues, then we know
\begin{equation}
\exp(\log(A(t)))=A(t)
\end{equation}
Take the derivative with respect to both sides we have
\begin{equation}
\exp(\log(A(t))) (\log(A(t)))'=A'(t)
\end{equation}
So we have
\begin{equation}
(\log(A(t)))'=A^{-1}(t)~A'(t)
\end{equation}
modulo convergence issues.
