By using the following definition:
$\delta ln(det(A))=ln(det(A+\delta A))-ln(det(A))$
I have to show that
$\delta ln(det(A))=tr(A^{-1}\delta A))+O(\delta A^{2})$
My initial approach was to set both equations equal to one another:
$ln(det(A+\delta A))-ln(det(A))=tr(A^{-1}\delta A))+O(\delta A^{2}) $
$ln(det(A)*det(1+A^{-1}\delta A))-ln(det(A))=tr(A^{-1}\delta A))+O(\delta A^{2})$
$ln(det(1+A^{-1}\delta A))=tr(A^{-1}\delta A))+O(\delta A^{2})$
Which is equivalent to
$det(1+A^{-1}\delta A)=e^{tr(A^{-1}\delta A)}+1$
From here on I am stuck and don't even know if my approach was right in the first place. I hope you can guide me towards the right direction.
You have to show that \begin{align} \ln(\det(I + A^{-1} \delta A)) &= \text{tr}(A^{-1}\delta A ) + \mathcal{O}(\delta A^2) \tag{$*$}. \end{align} Now, to denote the "smallness" of the matrix $\delta A$, I shall write it as $\epsilon B$, for some matrix $B$, and a small real parameter $\epsilon$ (afterall, that is the definition of the variation); so we now have to show that equality holds up to $\mathcal{O}(\epsilon^2)$.
Now, we have that $A^{-1} \cdot \delta A = \epsilon (A^{-1}B)$; let's regard this as an element of $M_{n \times n}(\Bbb{C})$. As such, the characteristic polynomial of $A^{-1}B$ splits, and thus it will be similar (over $\Bbb{C}$) to an upper-triangular matrix (if you'd like, you can take this to be the Jordan-Canonical form). That is, there exist $P, J \in M_{n \times n}(\Bbb{C})$, with $P$ invertible and $J$ upper-triangular such that \begin{align} A^{-1}B &= PJP^{-1}. \end{align} Therefore, by the multiplicative property of determinant, \begin{align} \ln(\det(I + \epsilon A^{-1}B)) &=\ln(\det(I + \epsilon PJP^{-1})) \\ &= \ln(\det(I + \epsilon J)) \tag{$\ddot{\smile}$}. \end{align} Now, the matrix $I + \epsilon J$ is upper-triangular, so the determinant is the product of diagonals, and it is easy to see that the determinant is $1 + \epsilon \cdot\text{tr}(J)$ to first order. But of course, the trace of a matrix is invariant under similarity transformations. Hence, \begin{align} \det(I + \epsilon J) &= 1 + \epsilon \cdot \text{tr}(J) + \mathcal{O}(\epsilon^2)= 1 + \epsilon \cdot \text{tr}(A^{-1}B) + \mathcal{O}(\epsilon^2). \end{align} Finally, we plug this back into $(\ddot{\smile})$, and by considering the Taylor expansion $\ln(1+x) = x + \mathcal{O}(x^2)$, it follows immediately that \begin{align} \ln(\det(I + \epsilon A^{-1}B)) &= \epsilon \cdot \text{tr}(A^{-1} B) + \mathcal{O}(\epsilon^2) \\ &= \text{tr}(A^{-1} \cdot \epsilon B) + \mathcal{O}(\epsilon^2) \end{align} In other words (apart from some notational difference), this proves the claim $(*)$.