It seems that most do not like Karl's answer. However I would like to stress that the suggestion of Karl is feasible, but one needs to be careful with the details.
Step 1. The function $Det\colon M_{n\times n}(\mathbb{C}) \mapsto \mathbb{C}$ is analytic.
This follows from the fact that $Det$ can be written as a polynomial function in terms of the coefficients of the matrices. In particular, its derivatives of arbitrary order exist and they are continuous, so the directional derivative
$$(A,B)\mapsto D_ADet\cdot B$$
is continuous in pairs of matrices $(A,B).$
Step 2. If $H$ is an invertible matrix then
$$D_{HA} Det \cdot B = (Det \ H)D_{A} Det \cdot H^{-1}B.\tag{I}$$
$$D_{AH^{-1}} Det \cdot B = (Det \ H^{-1})D_{A} Det \cdot BH.\tag{II}$$
$$D_{HAH^{-1}} Det \cdot B = D_{A} Det \cdot H^{-1}BH.\tag{III}$$
This follows from the definition of derivative and $Det(CH)=Det(HC)=Det(C)$.
Step 2 is especially interesting in the case of $A=Id$, since we get
$$D_{Id} Det \cdot B = D_{Id} Det \cdot \ H^{-1}BH.\tag{IV}$$
Step 3. (Complex) Diagonalizable matrices are dense in $M_{n\times n}(\mathbb{C})$.
This follows from a simple argument using perturbing the diagonal of the complex Jordan decomposition of a matrix. Diagonalizable matrices with complex values are dense in set of $n\times n$ complex matrices.
Step 4. If $B$ is a diagonal matrix we have
$$D_{I} Det\cdot B = trace \ B.$$
This follows from direct calculation more or less as Karl suggested. If the diagonal terms of $B$ are $b_{jj}= \lambda_j$, with $j\leq n$, then
$$\lim_{t\rightarrow 0} \frac{Det (Id+tB) - Det(Id)}{t}= \lim_{t\rightarrow 0} \frac{\prod_j (1+\lambda_j t) -1}{t}= \sum_j \lambda_j = trace \ B.$$
Step 5. If $H$ is invertible $trace \ H^{-1}BH = trace \ B.$
This follows from the fact that $trace (UV)=trace (VU)$ for every matrices $U$ and $V$, something one can prove by direct calculation since
$$trace (UV) = \sum_j \sum_k U_{jk}V_{kj} = \sum_k \sum_j V_{kj}U_{jk}=trace (VU),$$
so $trace \ H^{-1}BH = trace H^{-1}H B = trace \ B.$
Step 6. We have
$$D_{Id} Det\cdot B = trace \ B. \tag{V}$$
for diagonalizable matrices.
This follows from $(III)$ and Steps 4 and 5.
Step 6. We have $(V)$ for every matrix $B$.
Both sides of the equality are continuous with respect to $B$ and they coincide on diagonalizable matrices $B$. But this set is dense, so it follows that the equality holds everywhere.
Step 7. For every invertible matrix $A$ and a matrix $B$ we have
$$D_{A} Det\cdot B = Det(A) trace \ BA^{-1}.$$
Since $A$ is invertible, apply (II) with $H=A$ and replacing $B$ by $BA^{-1}$, and finally apply Step 6.
