Show that if $ A(t) \in \mathbb{R ^{n×n}}$ is a differentiable function of $t$ then $\frac{d(det A(t))}{dt}$ Let $\mathbb{R^{ n×n}}$ be the space of n × n real matrices. Show that if $ A(t) \in \mathbb{R ^{n×n}}$ is
a differentiable function of $t$ then $\frac{d(det A(t))}{dt}$ is the sum of the determinants of $n$ matrices, in which the $i$th matrix is $A(t)$ except that the $i$th row is differentiated. Use this result to prove that the directional derivative of the determinant function at the matrix $A \in \mathbb{R ^{n×n}}$ along the direction $B \in \mathbb{R^{ n×n}}$ is given by
$$(det)^{'} (A; B) = tr(Adj(A)B) = \langle Adj(A)^{T} , B \rangle $$
where $Adj(A)$ is the adjoint of $A$, and where the inner product on $\mathbb{R^{n×n}}$
is the trace inner product given by $\langle X, Y \rangle = tr(X^{T} Y )$
Conclude that
$$D(det)(A) = Adj(A) ^T $$.
 A: (After writing the answer, I noticed that the question mentioned
rows rather than columns. Since $\det A = \det A^T$ , it doesn't matter which we choose, so I will avoid the rewrite!)
Since $\det A$ is a polynomial in the entries of $A$ it is clear that
$\det$ is smooth, so we can compute the derivative by any means.
Let $[A]_{\bullet j}$ denote the $j$th column of $A$. We can write
$\det$ as a multilinear function of the columns $[A]_{\bullet 1},...,[A]_{\bullet n}$, that is, we have $\det A = M([A]_{\bullet 1},...,[A]_{\bullet n})$ for some multilinear $M$, (with
$(x_1,...,x_n) \mapsto M(x_1,...,x_n)$).
Since $M(x_1+h,...,x_n) = M(x_1,...,x_n) + M(h,...,x_n)$, we see that
${\partial M(x_1,...,x_n) \over \partial x_1}(h) = M(h,x_2,...,x_n)$,
and similarly for the other columns.
Now let $\phi(t) = \det A(t) = M([A]_{\bullet 1}(t),...,[A]_{\bullet n}(t))$, then using the chain rule, we have
\begin{eqnarray}
{d \phi(t) \over dt} &=& \sum_{k=1}^n {\partial M([A]_{\bullet 1}(t),...,[A]_{\bullet n}(t)) \over \partial x_k}( {d [A]_{\bullet k}(t) \over dt} )\\
&=& \sum_{k=1}^n M([A]_{\bullet 1}(t),...,{d [A]_{\bullet k}(t) \over dt},...,[A]_{\bullet n}(t))
\end{eqnarray}
Notice that $M([A]_{\bullet 1}(t),...,{d [A]_{\bullet k}(t) \over dt},...,[A]_{\bullet n}(t))$ is the determinant of $A$ with the
$k$th column replaced by ${d [A]_{\bullet k}(t) \over dt}$.
Consider the determinant of the matrix $A$ with the $k$th column replaced by the $b$.
Note that $M([A]_{\bullet 1},...,b,...,[A]_{\bullet n}) = \sum_{i=1}^n [b]_i M([A]_{\bullet 1},...,e_i,...,[A]_{\bullet n})$,
where $e_i$ is the $i$th unit vector.
We have $M([A]_{\bullet 1},...,e_i,...,[A]_{\bullet n}) = (-1)^{i+k}\det A_{ik}$, where $A_{ik}$ is the $(i,k)$ minor of $A$, and so
$M([A]_{\bullet 1},...,b,...,[A]_{\bullet n}) = \sum_{i=1}^n [b]_i (-1)^{(k+i)} \det A_{ik} = \sum_{k=1}^n [b]_i [\operatorname{adj} A]_{ki}$.
If $A(t) = A+tB$, then ${d [A]_{\bullet k}(t) \over dt} = [B]_{\bullet k}$, and we have
\begin{eqnarray}
{d \phi(t) \over dt} &=& \sum_{k=1}^n M([A]_{\bullet 1},...,[B]_{\bullet k},...,[A]_{\bullet n}) \\
&=& \sum_{k=1}^n \sum_{k=1}^n [B]_{ik} [\operatorname{adj} A]_{ki} \\
&=& \langle (\operatorname{adj} A)^T, B \rangle \\
&=& \operatorname{tr} ( (\operatorname{adj} A) B)
\end{eqnarray}
The last part of the question is a little ambiguous, as $D \det(A)$is a map $D \det(A): \mathbb{R}^{n \times n} \to \mathbb{R}^{n \times n}$ which doesn't have a convenient
'matrix' representation (that is, we cannot write the derivative
in the form $h \mapsto Ah$ for some matrix $A$).
The above shows that $d \det (A,B) = \langle (\operatorname{adj} A)^T, B \rangle$, where $d$ is the directional derivative.
If we let $\eta(t)=A+tB$, we have $d \det (A,B) = D (\det \circ \eta)(0) = D \det (A)(B) =  \langle (\operatorname{adj} A)^T, B \rangle$. If we choose $B=E_{ij}$ (the matrix of zeros except for
a one in the $i,j$ position), then we see that
$D \det (A)(E_{ij}) = [(\operatorname{adj} A)^T]_{ij}$. Presumably this
is the desired conclusion.
Alternative:
Here is a simpler derivation. As above, we note that $\det$ is
smooth, so we only need to compute the derivative, not show that
it exists. As above, we note that $D \det(A)(B) = d \det(A,B)$.
Suppose $A$ is invertible. Then $\det(A+tB) = \det A \det (I+ tA^{-1} B)$, so we have
$d \det(A,B) = D \det(A)(B) = (\det A) D \det(I)(A^{-1} B)$.
To compute $D \det(I)(H)$, note that
$D \det(I)(E_{ij}) = \delta_{ij}$ (where $\delta$ is the Kronecker
delta), and so
$D \det(I)(H) = \sum_i\sum_j [H]_{ij} D \det(I)(E_{ij}) = \sum_k [H]_{kk} = \operatorname{tr} H$.
Hence, if $A$ is invertible, 
$D \det(A)(B) = (\det A) \operatorname{tr} (A^{-1} B) = \operatorname{tr} ((\det A)A^{-1} B)= \operatorname{tr} ((\operatorname{adj}A) B)$.
If $A$ is not invertible, then let $A_n \to A$ where $A_n$
are invertible, then since $\operatorname{tr},\operatorname{adj}$
are continuous, we see that
$D \det(A)(B) = \lim_n D \det(A_n)(B) = \lim_n \operatorname{tr} ((\operatorname{adj}A_n) B) = \operatorname{tr} ((\operatorname{adj}A) B)$.
