Proof that $\frac{d}{dt} \log |A(t)| = \text{Tr}\left[A(t)^{-1} \frac{d}{dt} A \right]$ For a symmetric $n\times n$  matrix $A(t)$ that depends on the variable $t$, I want to prove that
$$
\frac{d}{dt} \log |A(t)| = \text{Tr}\left[A(t)^{-1} \frac{d}{dt} A(t) \right].
$$
Here, $|A|$ denotes the determinant of $A$ and $\text{Tr}[X]$ denotes the trace of $X$. How can I prove this equality?

The reason I ask this question: The equality is presented in equation C.22 of Pattern Recognition and Machine Learning of C. Bishop and the proof is left "as an exercise to the reader".

My attempt so far (to not clutter the notation, I leave out the dependency on $t$):
We know that $|A|=\prod_{i=1}^n \lambda_i$, where $\lambda_i$ denotes the $i$-th eigenvalue.
Hence, we can write
$$
\frac{d}{dt} \log|A|
=\frac{d}{dt}\log\prod_{i=1}^n\lambda_i
=\frac{d}{dt}\sum_{i=1}^n\log(\lambda_i)
=\sum_{i=1}^n\frac{d}{dt}\log(\lambda_i)
=\sum_{i=1}^n\frac{1}{\lambda_i} \frac{d}{dt}\lambda_i.
$$
The textbook further suggests to make use of the equalities
$$ 
A = \sum_{i=1}^n \lambda_i u_i u_i^T 
\quad \text{and} \quad 
A^{-1} = \sum_{i=1}^n \frac{1}{\lambda_i} u_i u_i^T, 
$$
where $u_i$ denotes the $i$-th eigenvector.
I do not know how to proceed from here. I also do not know if this is going into the right direction.
 A: Provided the problem is well-defined, i.e. matrix is diagonalizable, non-singular and determinant is positive, one should be able to use the orthonormality of eigenvectors, i.e.
$$
\mathbf{u}^T_i.\mathbf{u}_j=\sum_s u^\left(s\right)_i u^{\left(s\right)}_j=
\begin{cases}
1,\quad i=j \\
0,\quad otherwise
\end{cases}
$$
Where sum over $\dots^{\left(s\right)}$ is over individual components of the eigen-vectors.
Now:
$$
\begin{align}
\mbox{Tr}\left(\mathbf{A^{-1}}.\frac{d}{dt}\mathbf{A}\right)=&\mbox{Tr}\left(\left[\sum_i \mathbf{u}_i.\frac{1}{\lambda_i}.\mathbf{u}_i^T\right]\:\cdot\:\frac{d}{dt}\left[\sum_j \mathbf{u}_j .\lambda_j.\mathbf{u}_j^T\right]\right) \\
=&\sum_{sp}\sum_{ij}u^{(s)}_i\lambda_i^{-1}u^{(p)}_i\cdot\left(\frac{du^{(p)}_j}{dt}\lambda_ju^{(s)}_j+u^{(p)}_j\frac{d\lambda_j}{dt} u^{(s)}_j+u^{(p)}_j \lambda_j \frac{du^{(s)}_j}{dt}\right)\\
=&\sum_i \frac{1}{\lambda_i}\cdot\frac{d\lambda_i}{dt}+\sum_p\sum_i u^{(p)}_i\frac{du^{(p)}_i}{dt}+\sum_s\sum_i u^{(s)}_i\frac{du^{(s)}_i}{dt}\\
=&\sum_i \frac{1}{\lambda_i}\cdot\frac{d\lambda_i}{dt}+2\sum_s\sum_i u^{(s)}_i\frac{du^{(s)}_i}{dt}\\
=&\sum_i \frac{1}{\lambda_i}\cdot\frac{d\lambda_i}{dt}+\frac{d}{dt}\sum_s\sum_i u^{(s)}_i u^{(s)}_i \\
=&\sum_i \frac{1}{\lambda_i}\cdot\frac{d\lambda_i}{dt}
\end{align} 
$$
Since
$$
\sum_i \sum_s u^{(s)}_i u^{(s)}_i = \sum_i 1=\mbox{number of components}=\mbox{const}
$$
Which, I think, was what you were after
A: $\newcommand{\Tr}{\mathrm{Tr}}$
$\newcommand{\adj}{\mathrm{adj}}$
In fact, the equality holds as long as $\det(A(t)) > 0$ (the symmetry condition is not needed).
The key identity we need is that for any $A(t)$ ($\det(A(t))$ may equal to $0$):
\begin{align*}
\frac{d(\det(A(t)))}{dt} = \sum_{i = 1}^n\sum_{j = 1}^n \frac{d a_{ij}(t)}{dt}A_{ij}(t) = \Tr\left(\frac{d(A(t))}{dt}\mathrm{adj}(A(t))\right), \tag{1}
\end{align*}
where
\begin{align*}
\adj(A(t)) = \begin{pmatrix} A_{11}(t) & A_{21}(t) & \cdots & A_{n1}(t) \\
A_{12}(t) & A_{22}(t) & \cdots & A_{n2}(t) \\
\vdots & \vdots & \ddots & \vdots \\
A_{1n}(t) & A_{2n}(t) & \cdots & A_{nn}(t)
\end{pmatrix}
\end{align*}
is the adjugate matrix of $A(t)$ thus it satisfies
\begin{align*}
A(t)\adj(A(t)) = \det(A(t))I_{(n)}.
\end{align*}
When $\det(A(t)) > 0$ hence $A(t)$ is invertible, $\adj(A(t)) = \det(A(t))A(t)^{-1}$, whence by the chain rule we have
\begin{align*}
\frac{d\log(\det(A(t)))}{dt} &= \frac{1}{\det(A(t))}\frac{d(\det(A(t)))}{dt}
= \frac{1}{\det(A(t))}\Tr\left(\frac{d(A(t))}{dt}\mathrm{adj}(A(t))\right) \\
&= \frac{1}{\det(A(t))}\Tr\left(\frac{d(A(t))}{dt}\det(A(t))A(t)^{-1}\right) 
 = \Tr\left(\frac{d(A(t))}{dt}A(t)^{-1}\right).
\end{align*}
The proof of the first equality in $(1)$ is subtle and quite lengthy. If you are interested, I may post it as an appendix.
A: Suppose that there is $B(t)$ such that
$$ A(t)=e^{B(t)}. $$
Then
$$ \frac{d}{dt}\log |A(t)|=\frac{d}{dt}\log |e^{B(t)}|=\frac{d}{dt}\log e^{\text{Tr}B(t)}=\frac{d}{dt}\text{Tr}B(t). $$
On the other hand,
$$ \text{Tr}\left[A(t)^{-1} \frac{d}{dt} A(t) \right]=\text{Tr}\bigg[e^{-B(t)}e^{B(t)}B'(t)\bigg]=\text{Tr}B'(t)=\frac{d}{dt}\text{Tr}B(t). $$
So
$$ \frac{d}{dt}\log |A(t)|=\text{Tr}\left[A(t)^{-1} \frac{d}{dt} A(t) \right].$$
A: Nicholas Higham published a wonderful paper on the Unwinding Function
$$\eqalign{
{\cal U}(z) &= \frac{z-\log(e^z)}{2\pi i} 
  = \left\lceil\frac{Im(z)-\pi}{2\pi}\right\rceil \;\in\;{\mathbb Z} \\
}$$
Using this function one can write the following identity
$$\eqalign{
\log\det A
 &= {\rm tr}\log A - 2\pi i\cdot {\cal U}({\rm tr}\log A) \\
 &= {\rm tr}\log A - 2n\pi i \\
}$$
One of the many interesting properties of the unwinding function is that its derivative is zero, which leads directly to your formula, i.e.
$$\eqalign{
\frac{d(\log\det A)}{dt}
 &= \frac{d({\rm tr}\log A)}{dt} \\
 &= \operatorname{tr}\left(A^{-1}\frac{dA}{dt}\right) \\
}$$
The final equality follows from the identity
$$\eqalign{
\frac{d({\rm tr}(f(A)))}{dt}
 &= \operatorname{tr}\left(f'(A)\cdot\frac{dA}{dt}\right) \\
}$$
where $f$ is any analytic function and $f'$ is its derivative.
A: From $A(t)$ diagonalizable and $tr(BC)=tr(CB)$ we have $\log \det A(t)=tr \log A(t)$.
Next, $(A(t)^n)'=\sum_{k=0}^{n-1} A^k A'(t)A^{n-1-k}$ gives that for $f(z)=\sum_{n\ge 0} c_n z^n$ analytic on a disk containing $\|A(t)\|$ then $$tr(f(A(t)))'=\sum_{n=0}^\infty c_n \sum_{k=0}^{n-1} tr(A^k A'(t)A^{n-1-k})=\sum_{n=0}^\infty c_n \sum_{k=0}^{n-1} tr(A^{n-1}A'(t))$$ $$=tr( f'(A(t)) A'(t))$$
$f(z)=\log z$ is not analytic on such a disk, but by analytic continuation in $s$ of $tr(f(I+s(A(t)-I)))'$ it will still hold, ie. $$tr(\log(A(t)))' = tr(A(t)^{-1}A'(t))$$
No need that the eigenvalues of $A(t)$ are positive, just that $A(t)$ is differentiable, $\det(A(t))\ne 0$ and our choice of $tr \log A(t)$ is continuous in $t$.
