A proof of: The derivative of the determinant is the trace I want to solve the following problem:

Show that the derivative of $\mbox{det}:GL(n,\mathbb{R})\rightarrow\mathbb{R}$
    at $I\in GL(n,\mathbb{R})$
    is given by 
  $$\mbox{det}_{*}(I)(X)=\mbox{tr}X$$

I would like you to check my proof, and answer the question in the end.
My Attempt: I'll denote $N=GL(n,\mathbb{R})$
 . Also, $\simeq$
  will be used for vector space isomorphisms and $\cong$
  will be used for diffeomorphisms. 
We know that $\mbox{det}_{*}(I):T_{I}N\rightarrow T_{det(I)=1}\mathbb{R}$
 .
Let $X\in T_{I}N$
 . We can write $X$
  in a basis of $T_{I}N$
 . So let us find a basis $of T_{I}N$
 : we know that $T_{I}N\simeq M_{n}(\mathbb{R})$
 , so we can get a basis of $T_{I}N$
  from a basis of $M_{n}(\mathbb{R})$
  using an isomorphism. The function $$f:T_{I}N \rightarrow M_{n}(\mathbb{R}) \\
[\gamma] \mapsto \gamma'(0)$$
is known to be an isomorphism. Furthermore, ${E_{ij}}$
  is a basis for $M_{n}(\mathbb{R})$
 , where $E_{ij}$
  is the $n\times n$
  matrix whose entries are all zero except the entry $i,j$
 , which is $1$
 . Thus, a basis for $T_{I}N$
  is ${f^{-1}(E_{ij})}$
 . Now, $f^{-1}(E_{ij})$
  is the equivalence class of curves $\gamma:\mathbb{R}\rightarrow N$
  such that $\gamma(0)=I$
  and $\gamma'(0)=E_{ij}$
 . Hence, a representative of this equivalence class is $\alpha_{ij}(t)=I+tE_{ij}$
 , and so we can write ${f^{-1}(E_{ij})}={[\alpha_{ij}]}$
 .
Hence, we can write $X=\overset{n}{\underset{i,j=1}{\sum}}x_{ij}[\alpha_{ij}]$
 .
Let us see how $det_{*}$
  acts on the basis elements $[\alpha_{ij}]$
 . 
We have $\mbox{det}_{*}(I)([\alpha_{ij}])=[\mbox{det}\circ\alpha_{ij}]_{1}$
by definition of derivative (the subscript 1
  reminds us that the equivalence relation of this equivalence class is different, since it is defined on the set of all curves of the type $\gamma:\mathbb{R}\rightarrow\mathbb{R}$
  such that $\gamma(0)=\mbox{det}(I)=1
 ).$
Now, $\mbox{det}\circ\alpha_{ij}:\mathbb{R}\rightarrow\mathbb{R}$
  is such that $$\mbox{det}\circ\alpha_{ij}=\mbox{det}(\alpha_{ij}(t))=\mbox{det}\left(I+tE_{ij}\right)=\mbox{det}\left(\left[\begin{array}{ccc}
1 &  & \mathbb{O}\\
 & \ddots\\
\mathbb{O} &  & 1
\end{array}\right]+\left[\begin{array}{cccc}
\mathbb{O} &  &  & \mathbb{O}\\
 &  & t\,(i,j\mbox{ entry})\\
\\
\mathbb{O} &  &  & \mathbb{O}
\end{array}\right]\right)$$
 . The matrix is triangular (or simply diagonal), and so the determinant is the product of the diagonal elements. Hence, $\mbox{det}\circ\alpha_{ij}=1+t\delta_{ij}$
 , with $\delta_{ij}$
  the Kronecker delta.
Hence, $$\mbox{det}_{*}(I)([\alpha_{ij}])=[1+t\delta_{ij}]_{1}\in T_{1}\mathbb{R}$$
Finally, $$\mbox{det}_{*}(I)(X)=\overset{n}{\underset{i,j=1}{\sum}}x_{ij}\mbox{det}_{*}(I)([\alpha_{ij}])=\overset{n}{\underset{i,j=1}{\sum}}x_{ij}[1+t\delta_{ij}]_{1}$$
Now, I noticed that, if I for some reason use the isomorphism $$g:T_{1}\mathbb{R} \rightarrow \mathbb{R} \\
[\gamma]_{1} \mapsto \gamma'(0)$$
to “identify” $\alpha_{ij}$
  with $g(\alpha_{ij})=\delta_{ij}$
  and use that instead of $[1+t\delta_{ij}]_{1}$
 , I get $\overset{n}{\underset{i,j=1}{\sum}}x_{ij}\delta_{ij}=\mbox{tr}X$
 .
My question is: why is this last step (since "Now, I noticed...") legitimate?
 A: Since it seems you want to understand all the various identifications involved, let me introduce some notation to try and clarify what is going on.
Let $V$ be a finite dimensional real vector space (endowed with the natural smooth structure) and let $U \subseteq V$ be an open subset. Assume we are given a smooth function $F \colon U \rightarrow \mathbb{R}$. Then we can calculate three a priori distinct things:


*

*We can calculate the standard multivariable directional derivative of $F$ at a point $p \in U$ in the direction $v \in V$. I'll denote it by
$$ DF|_{p}(v) := \lim_{t \to 0} \frac{F(p + tv) - F(p)}{t}. $$
Note that $DF|_{p} \colon V \rightarrow \mathbb{R}$.

*We can treat $U$ as a smooth manifold and calculate the differential of $F$ at a point $p \in U$. This will give us a map $dF|_{p} \colon T_p U \rightarrow \mathbb{R}$.

*We can treat both $U$ and $\mathbb{R}$ as smooth manifolds and calculate the "full differential" of $F$ at a point $p$ which I'll denote by $F_{*}$. This will give us a map $F_{*}|_{p} \colon T_p U \rightarrow T_{F(p)} (\mathbb{R})$.


What is the relation between the three distinct maps? Like you noted, we have natural isomorphisms $f \colon T_pU \rightarrow V$ and $g \colon T_{F(p)}(\mathbb{R}) \rightarrow \mathbb{R}$. In terms of those isomorphisms, we have the relations
$$ dF|_{p} = DF|_{p} \circ f, \,\,\, dF|_{p} = g \circ F_{*}|_{p}, \,\,\, g^{-1} \circ DF|_{p} \circ f = F_{*}|_{p}. $$
Now, in your case, we have $F = \det$ and $p = I$. By the formula you are given, it seems that you are asked to calculate $DF|_{p}$. Instead, you have tried to calculate $F_{*}|_{p}$, hence the need for isomorphisms $f$ and $g$ to relate the formula you are asked to show with your calculation. Note that like Qiaochu said in his comment, it is much easier to calculate $DF|_{p}(X)$ without choosing a basis but your more complicated calculation agrees with the expected result.
A: Another approach uses standard coordinate : note that, as a function of several variables, the determinant is particularly simple as it is linear in each coordinate.
If you developp the determinant using standard rule, you see that is $X=(x_{i,j})$ is a matrix and the index $i$ is fixed $\det M= \sum _{j=1}^n x_{i,j} \det X_{i,j} (-1)^{i+j}$, where $X_{i,j}$ is obtained by erasing the i-th raw and column of $X$.
If follows that $({ \partial  \over \partial x_{i,j}} \det ) X= (-1)^{i+j} \det X_{i,j}$
In particular if $X= Id$ is the identity matrix, $({ \partial   \over \partial x_{i,j}} \det ) Id=0$ if $i\not = j$ and $({ \partial   \over \partial x_{i,i}} \det ) Id=1$
Whence $d \det ({Id}) M= \sum _{i,j} ({ \partial   \over \partial x_{i,i}} \det )(Id) m_{i,j} =\sum _i m_{i,i}$
This approach immediately gives you the gradient of the determinant at any point $X$ : $\vec {grad} (\det )X$ is the matrix $(-1)^{i+j} \det X_{i,j}$
