Diagonalization of a matrix $A \in GL(n,\mathbb{C})$ I am trying to show that 
$$
\frac{d}{dt} \log \det A_t = Tr (A^{-1}_tA'_t)
$$
where $A_t \in GL(n,\mathbb{C})$ and $A'_t = \frac{d}{dt}A_t$.
I think I can show why this is the case if $A$ is diagonalizable,but I am not sure how to do it in case it is not. 
Does $GL(n,\mathbb{C})$ have a property that allows me to deduce the above equality even if the matrix is not diagonalizable ? 
Thanks a lot in advance for any hints.
 A: It needs to prove :
$$\frac{1}{|A_t|}\frac{d}{dt}|A_t|=tr(A_t^{-1}A_t')$$
namely ,$$\frac{d}{dt}|A_t|=tr(A_t^{*}A_t')$$
the adjoint matrix
$$A^{*}=\left( \begin{array}{cccc}
                               A_{11}^{'}(t) &  A_{21}^{'}(t) & \cdots &  A_{n1}^{'}(t)\\
                               A_{12}^{'}(t) &  A_{22}^{'}(t) & \cdots &  A_{n2}^{'}(t) \\
                               \cdots & \cdots & \cdots & \cdots \\
                                A_{1n}^{'}(t) &  A_{2n}^{'}(t) & \cdots &  A_{nn}^{'}(t) \\
                             \end{array}
                           \right)$$
and $$A^{'}=\left( \begin{array}{cccc}
                               a_{11}^{'}(t) &  a_{12}^{'}(t) & \cdots &  a_{1n}^{'}(t)\\
                               a_{21}^{'}(t) &  a_{22}^{'}(t) & \cdots &  a_{2n}^{'}(t) \\
                               \cdots & \cdots & \cdots & \cdots \\
                                a_{n1}^{'}(t) &  a_{n2}^{'}(t) & \cdots &  a_{nn}^{'}(t) \\
                             \end{array}
                           \right)$$
so $$tr(A_t^{*}A_t')=\sum_{j=1}^{n}\sum_{i=1}^{n} A_{ij}(t)a_{ij}^{'}(t)$$
Meanwhile,
$$\frac{d}{dt}|A_t|=\sum_{j=1}^{n}\left|\begin{array}{ccccc}
                               a_{11}(t) &  a_{12}(t) & \cdots &  a_{1n}(t)\\
                               \cdots & \cdots & \cdots & \cdots \\
                               a_{j1}^{'}(t) &  a_{j2}^{'}(t) & \cdots &  a_{jn}^{'}(t) \\
                               \cdots & \cdots & \cdots & \cdots \\
                                a_{n1}(t) &  a_{n2}(t) & \cdots &  a_{nn}(t) \\
                                        \end{array}\right|=\sum_{j=1}^{n}\sum_{i=1}^{n}a_{ji}^{'}(t) A_{ji}(t)$$
So, this proposition is already proven.
