Jacobian and continuity Trying to prove that if $g:\mathbb R^n \to \mathbb R^n$ is continuously differentiable then $\det J(g(x))$ is continuous. It seems obvious as $\det J(g(x))$ is a polymnomial consisting of the partial derivatives of $g$ which all are continuous, but I am having troubles proving it explicity. I would like to do with via epsilon delta proof
 A: As you said, the fact that $\det$ is a polynomial in the entries proves what you need. You are probably using the definition that a function is $C^1$ if its partial derivatives are continuous. Hence, defining $\partial: \mathbb{R}^n \rightarrow \mathbb{R}^{n^2}$ as $\partial(x)=(\ldots, \frac{\partial g_i}{\partial x_j},\ldots)$, we see (by hypothesis) that $\partial$ is continuous. Now, it is easy to see that there is a function $\overline{\det
}$ which is a sum of multiplication of projections that yields ${\overline{\det}}\circ \partial=\det(g(x))$. Since sums are continuous, projections are continuous and multiplications are continuous, we get that ${\overline{\det}} \circ \partial$ is continuous, hence so is $\det (g(x))$.
Also, you could argue as follows: a function $f: \mathbb{R
^n} \rightarrow \mathbb{R}^n$ is $C^1$ if the derivative 
$$Df: \mathbb{R}^n \rightarrow L(\mathbb{R}^n, \mathbb{R}^n) $$
$$ x \mapsto D_xf$$
is continuous. Since $\det : L(\mathbb{R}^n, \mathbb{R}^n) \rightarrow \mathbb{R}$ is continuous and composition of continuous functions is continuous, the result follows. The advantage of this approach is that it is coordinate-free (although you will probably use coordinates to arrive at the fact that $\det$ is continuous). 
