Derivative of determinant at some point Let $c:\mathbb{R} \rightarrow \mathbb{M}_n(\mathbb{R})$ defined by $$c(t)=A e^{tB}$$ where $A\in GL(n,\mathbb{R})$ and $B \in \mathbb{M}_n(\mathbb{R})$.
The question ask me to find $c'(0)$ and $(\text{det}\circ c)'(0)$.
Then find the derivative of $\text{det}:\mathbb{M}_n(\mathbb{R})\rightarrow \mathbb{R}$ at $A$

I did the first part, where:
$$c'(0)=AB$$
$$(\text{det}\circ c)'(0)=\text{det}(A)  \text{Tr}(B)$$
But I don't know how to deduce $D\text{det}_A$, the derivative of $\text{det}$ at $A$.
Here is what I was thinking, since $D\text{det}_A$ takes initial tangent vector (which is $c'(0)$) to initial tangent vector of $\text{det}\circ c$. 
Hence, 
$$D\text{det}_A(c'(0)) =(\text{det}\circ c)'(0) $$
and I don't know what to do next.
 A: The idea was to exploit the fact that $$\det A\, tr B=(\det\circ c)'(0) = \nabla_X(\det X)'\big|_{X=c(0)}:c'(0)$$ with $\nabla_X(\det X)'$ being the gradient   of the determinant with respect to coefficients of the matrix.
Hence, we get
$$\det A\,tr B = \nabla (\det A):(AB)$$
As $\det A\ne0$, we can chose $AB$ to be whatever matrix we want: For example, if $AB=e_{ij}$, then $(\nabla (\det A):(AB)_{ij}=\nabla (\det A)_{ij}$, and on the other hand we have $B=A^{-1}e_{ij}$, therefore $tr B = (A^{-1})_{ji}$.
Finally, by enumerating all possible indices $i,j$ we conclude that
$$\nabla(\det A) = \det A(A^{-1})^T.$$
As a side note, I think that Laplace development for determinant together with the notion of comatrix allow to obtain the above formula in a much easier way.
An addendum: where the double contraction comes from: let $X $ with components $x_{ij}$ be a square matrix  and we study the gradient of $f(X)$ - certain numerical function depending on $X$. This gradient $\nabla_X f(X)$ is a matrix whose components are $$(\nabla_X f(X))_{ij}=\frac{\partial f(X)}{\partial x_{ij}}.$$
Now let us suppose that $X=X(t)$ and we want the derivative of the function $t\to f(X(t))$. By the rule of composite derivative we get
$$\frac{d}{dt}f(X(t)) = \sum_{ij}\frac{\partial f(X)}{\partial x_{ij}}\big|_{X=X(t)}\cdot \frac{d}{dt}x_{ij}(t). $$
Now, take into account that $\frac{d}{dt}x_{ij}(t)$ form a matrix $X'(t)$ and the definition of double contraction: $A:B = \sum_{i,j}A_{ij}B_{ij}$ to obtain
$$\frac{d}{dt}f(X(t)) = \nabla_X f(X(t)):X'(t).$$
The derivative of the matrix is taken component-wise (i.e. "the derivative of a matrix is a matrix of derivatives").
A: For finding $d(\det)_A$, let's use what you've already computed.  Given $B\in M_n(\mathbb{R})$, we have the curve $c(t)=A\exp{tA^{-1}B}$ with $c(0)=A$ and $c'(0)=B$.  Thus
\begin{align*}
d(\det)_A(B)&=d(\det)_A(c'(0))\\
&=(\det\circ c)'(0)\\
&=\det{A}\operatorname{tr}(A^{-1}B).
\end{align*}
