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][1] 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][2].)


  [1]: http://link.springer.com/article/10.1007/s10659-008-9169-x
  [2]: https://drive.google.com/open?id=0BwXVVor8UZUGLVNDYmZQcDlmY2c