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I want to show that 1 is a regular value of $\det:GL(n,\mathbb{R})\rightarrow\mathbb{R}$, i.e. that for all $A\in GL(n,\mathbb{R})$ the rank of $D(f\circ\phi^{-1})(\phi(A)):\mathbb{R}^{n^{2}}\rightarrow\mathbb{R}$ is maximal (and hence equal to 1), where $\phi$ is a chart on $GL(n,\mathbb{R})$ at $A$.

I'm guessing I should prove that $D(f\circ\phi^{-1})(\phi(A))$ is surjective, but I don't know how to do this because I don't have an expression for $D(f\circ\phi^{-1})(\phi(A))$. I do know that $\mbox{det}_{*}(I)(X)=\text{tr}(x)$ but I don't know how to go from here.

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    User Braindead's answer here http://math.stackexchange.com/questions/1712667/show-that-sln-mathbbr-is-a-n2-1-smooth-submanifold-of-mn-mathbb shows any nonzero number is a regular value of $\det$. – 2017-02-27

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We can use the formula we have for computing the determinant by expanding it along any row or column, namely if we expand along the $i$-th row we obtain:

$det(X) = \sum_{j=1}^n(-1)^{i+j}a_{ij}m_{ij}$

Where $m_{ij}$ is the determinant of the $i,j$-th minor of $X$. We can see that the partial with respect to the $i,j$-th entry will then be $\frac{\partial det}{\partial a_{ij}} = (-1)^{i+j}m_{ij}$. Now all that is left to check is that if the matrix has a nonzero determinant, the jacobian will have rank at least 1. Can you take it from here?

A cool fact, you can use this along with the regular value theorem to prove that $SL_n(\mathbf{R})$ is a Lie subgroup $GL_n(\mathbf{R})$!

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    let me try and see if I understood: The rank of $f=\det$ at $A$ is, by definition, the rank of the linear map $D(f\circ\phi^{-1})(\phi(A)):\mathbb{R}^{n^{2}}\rightarrow\mathbb{R}$, where $\phi$ is a chart on $GL(n,\mathbb{R})$ at $A$. We have $\det_{\bullet}(X):=\det\circ\phi^{-1}(X)=\overset{n}{\underset{j=1}{\sum}}(-1)^{i+j}a_{ij}m_{ij}$. Now, to $D\det_{\bullet}(\phi(A))$ corresponds the Jacobian $J(\det_{\bullet})|_{\phi(A)}$ whose entries are $\frac{\partial\det_{\bullet}}{\partial a_{ij}}(\phi(A))=(\overset{n}{\underset{j=1}{\sum}}(-1)^{i+j}m_{ij})(\phi(A))$. – 2017-02-27
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    There is something wrong here: $\frac{\partial\det_{\bullet}}{\partial a_{ij}}$ should be a function of real numbers, not a real number itself... – 2017-02-27