Relation of Determinant and trace of a matrix concerning derivative If there is a function 
$F(t) = det(I_{n} + tA)$
where 
$A$ is an $n \times n$ matrix,
$t$ is an arbitrary real number, 
and $I_{n}$ is $n \times n$ identity matrix,
is it true that the derivative of $F(t)$ at
$t = 0$ is equal to the trace of $A$? 
That is, 
$F'(0) = Tr(A)$
I currently know that the trace is the sum of the diagonal entries of a matrix but I am not sure how I should go about differentiating the right hand side. 
Is there a general formula for finding determinant that I could possibly differentiate? 
It seems like there is something called 'big formula for determinant' but I am not sure how that can be used for this problem.
Any help would be appreciated.
 A: The formula for the detrminant of a mattrix is 
$$\det(B)=\sum_{\sigma \in S_n} \epsilon(\sigma) b_{1 \sigma(1)} \cdot ... \cdot b_{n \sigma(n)} $$
Now, in your case 
$$b_{i,j}= \left\{ 
\begin{array}{lc}
1+ta_{i,i} & \mbox{if i=j} \\
ta_{i,j} &\mbox{ if } i \neq j 
\end{array}
\right.$$
Now, split your formula in 2 terms
$$\det(F(t))=b_{11}b_{22}\cdot...\cdot b_{nn}+\sum_{\sigma \in S_n, \sigma \neq e} \epsilon(\sigma) b_{1 \sigma(1)} \cdot ... \cdot b_{n \sigma(n)}$$
Now, each term in $Q(t):=\sum_{\sigma \in S_n, \sigma \neq e} \epsilon(\sigma) b_{1 \sigma(1)} \cdot ... \cdot b_{n \sigma(n)}$ contains two non-diagonal entries.
Since $Q(t)$ is a polynomial, and each non-dagonal entry has a factor of $t$, it follows that $t^2$ is a factor of $Q(t)$ and hence 
$$Q'(0)=0$$
Next, let 
$$P(t):=b_{11}b_{22}\cdot...\cdot b_{nn}=(1+ta_{11})(1+ta_{22})...(1+ta_{nn})$$
By using the product rule you immediatelly get 
$$P'(0)=a_{11}+a_{22}+..+a_{nn}=tr(A)$$
Since $\det(F(t))=P(t)+Q(t)$ you get
$$(\det(F(t)))'|_{t=0}=P'(0)+Q'(0)=tr(A)$$
A: Let $B$ be a curve of isomorphisms in a vector space $V$, with $\dim V=n$. Then if $\omega$ is a volume form on $V$, and $v_1,...,v_n\in V$ we have that $$\begin{align}(\det (B))^\cdot& \omega(v_1,...,v_n)=\big(\det( B)\omega(v_1,...,v_n)\big)^\cdot\\ &= (\omega(Bv_1,...,Bv_n))^\cdot \\ &=\sum_{i=1}^n \omega(Bv_1,...,\dot{B}v_i,...,Bv_n)\\ &=\sum_{i=1}^n \omega(Bv_1,...,BB^{-1}\dot{B}v_i,...,Bv_n)\\ &= \det (B )\sum_{i=1}^n \omega(v_1,...,B^{-1}\dot{B}v_i,...,v_n) \\ &= \det (B ){\rm tr}(B^{-1}\dot{B})\omega(v_1,...,v_n).\end{align}$$By arbitrariety of the vectors and of the volume form chosen, it follows that $(\det (B))^\cdot = \det (B) {\rm tr}(B^{-1}\dot{B})$. If you don't like the dot notation and want to make the dependence on a parameter $t$ explicit, this means that $$\frac{{\rm d}}{{\rm d}t} \det B(t) = \det (B(t)){\rm tr}(B(t)^{-1}B'(t)).$$
Now consider $F(t)=\det({\rm Id}+tA)$. For small enough $t$, the matrix ${\rm Id}+tA$ is non-singular, so we may use the above to compute the derivative at $t=0$ as $$F'(0) = \det({\rm Id}) {\rm tr}({\rm Id}^{-1}A)={\rm tr}(A),$$as wanted.
A: You have $\det(\lambda I-A) = (\lambda-\lambda_1)\cdots(\lambda-\lambda_n)$, where $\lambda_1,\ldots,\lambda_n$ are the (possibly complex) eigenvalues of $A$. Hence,
\begin{align}
\det(I+tA)
&= \det((-t)(-t^{-1}I-A)) = (-t)^n\det(-t^{-1}I-A)\\
&= (-t)^n(-t^{-1}-\lambda_1)\cdots(-t^{-1}-\lambda_n)\\
&= (1+t\lambda_1)\cdots(1+t\lambda_n)\\
&= 1+(\lambda_1+\ldots+\lambda_n)t+a_2t^2+\ldots+a_nt^n.
\end{align}
Now, deriving with respect to $t$ and setting $t=0$ yields the trace of $A$.
A: Here is a general rule for the differential of the determinant of $X$.
$$d\big(\det(X)\big) = \det(X)\;{\rm Tr}(X^{-1}dX)$$
In the current problem we have
$$\eqalign{
&X = (I + At) \;\implies\; dX = A\,dt \\
&d\big(\det(X)\big) = \det(X)\;{\rm Tr}\big(X^{-1}A\,dt\big) \\
&\frac{d\big(\det(X)\big)}{dt} = \det(X)\;{\rm Tr}\big(X^{-1}A\big) \\
}$$
Then as $t\to 0$
$$\eqalign{
X &\to I \\
\det(X) &\to \det(I) = {\tt 1} \\
{\rm Tr}(X^{-1}A) &\to {\rm Tr}(A) \\
\\
}$$
