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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 eigenvalues 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.)

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.)

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 $$ \log'(A(t))_{ab;cd} = - (A(t)^{-1})_{ac} (A(t)^{-1})_{db} $$ 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 eigenvalues 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.)

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 eigenvalues 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.)

First, note that 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 $$ \log'(A(t))_{ab;cd} = - (A(t)^{-1})_{ac} (A(t)^{-1})_{db} $$ 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 eigenvalues 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.)

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 $$ \log'(A(t))_{ab;cd} = - (A(t)^{-1})_{ac} (A(t)^{-1})_{db} $$ 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 eigenvalues 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.)