Functions and determinants Suppose that $f : W \rightarrow \mathbb R^n$ is a continuously differentiable mapping defined in an open neightbourhood $W \in \mathbb R^n$ of the origin such that $f(0)=0$ and $f'(0)=I$, the identity matrix.
Is it always possible to chose a closed cell in $W$ such that $det{f'(x)} \ne 0$ for all $x$ in the closed cell?
 A: $f \in \mathscr{C}'$ means that $x \rightarrow f'(x)$ is a continuous mapping from $W$ to $\mathscr{L}\left ( \Bbb{R}^n,\Bbb{R}^n \right )$. In particular: 
$$\left (\forall \varepsilon >0 \right )\left ( \exists \delta >0\right )\left (\left |x-y \right |<\delta \Rightarrow \left \| f'(0)-f'(y)\right \|<\varepsilon \right ), \ y \in W$$
($\left \| \ . \right \|$ being the matrix norm induced by the Euclidean norm)
Recall the following result: 
Let $\Omega$ be the set of all invertible operators in $\mathscr{L}\left ( \Bbb{R}^n,\Bbb{R}^n \right )$. Suppose $A \in \Omega$ and $B \in \mathscr{L}\left ( \Bbb{R}^n,\Bbb{R}^n \right )$ with $\left \| A^{-1} \right \| \left \| A-B \right \|<1$, then $B \in \Omega$. 
Take $\delta >0$ such that the neighbourhood of $0$ of radius $\delta$ is contained in $W$ ($W$ is open) and such that $\left \| I - f'(y) \right \|<1/2$ for all y with $\left | y \right |<\delta$ 
Since $f'(0)=I$ we have $\left \| \left (f'(0) \right )^{-1} \right \|=\left \| I \right \|=1$. Therefore: 
$$\left \| \left (f'(0) \right )^{-1} \right \|\left \| I-f'(y) \right \|<1/2<1 \ , \ \left | y \right | < \delta $$
It then follows that $f'(y)$ is invertible for all $y$ such that $\left |y\right |<\delta$ and hence $\det f'(y) \neq 0$.
All that's left to do is choose some closed cell which is contained in the neighbourhood of radius $\delta$ of $0$. This is possible since said neighbourhood is open, I leave the details to you. 
EDIT: Recall that given a matrix $A$, its determinant is a polynomial of its entries, and hence it is continuous as a function of the entries of $A$ (one can also prove continuity by induction since the determinant can be computed from the minors). Therefore, if one views $\det f'(x)$ as the composition $x \rightarrow f'(x) \rightarrow \det f'(x)$ this is a function $\Bbb{R} \rightarrow \Bbb{R}$ and it is continuous since the composition of continuous functions is continuous. If we call this function say $g(x)=\det(f'(x))$ we have:
$g(0)=1$ and by continuity $\exists \delta >0 \ , \  |x|<\delta $ implies $\left |1-g(x) \right |<1/2$
Hence $g(x)>1/2$ for $|x|<\delta$. In particular, $g(x)\neq 0$ for such $x$. 
