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As far as I understand, the rank of a map $f:GL(n,\mathbb{R})\rightarrow\mathbb{R}$ at $A\in GL(n,\mathbb{R})$ is 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 $A$.

However, in Show that $SL(n, \mathbb{R})$ is a $(n^2 -1)$ smooth submanifold of $M(n,\mathbb{R})$, user Ben West wrote a beautifully simple proof where he uses that

$$d(\det)_A(A)=\lim_{t\to 0}\frac{\det(A+tA)-\det(A)}{t}$$

and that if the rank of $d(\det)_A$ is the rank of $f$.

I'm not seeing the connection between the two "definitions". Probably because I don't really get what $d(\det)_A$ is: it is not the derivative $\det_*$ which is a map between the tangent spaces, but it is not the real-space-derivative $D(f\circ\phi^{-1})(\phi(A)):\mathbb{R}^{n^{2}}\rightarrow\mathbb{R}$.

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    Since $\mathrm{GL}(n, \Bbb R)$ is an open subset of the space $M(n, \Bbb R)$ of $n \times n$ matrices, which can in turn identify with $\Bbb R^{n^2}$, there is a canonical choice of $\phi$, namely the inclusion $\mathrm{GL}(n, \Bbb R) \hookrightarrow \Bbb R^{n^2}$, which we can regard as the restriction of the identity map. For this $\phi$, we thus have, $D(f \circ \phi^{-1})(\phi(A)) = Df(A)$. – 2017-02-27
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    On the other hand, $d(\det)_A$ *is* the derivative, but the formula you cite is implicitly using the identifications $T_A \textrm{GL}(n, \Bbb R) \cong T_A \Bbb R^{n^2} \leftrightarrow \Bbb R^{n^2}$ and $T_{\det A} \Bbb R \leftrightarrow \Bbb R$. – 2017-02-27
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    @Travis Still, we want to find the rank of $Df(A)$, not the rank of the derivative $\det_*(A)$ – 2017-02-27
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    For $f = \det$ these two are simply two notations for the same quantity (again, under the identifications I mention). – 2017-02-27

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