I've recently been taking 18.02sc Multivariable Calculus on MIT OpenCourseWare, which states the following in one of their course notes:
$$M \hat i + N \hat j = \nabla f \implies M dx + N dy \text{ is exact}$$
This makes sense to me, as:
$$ \nabla f = M \hat i + N \hat j \implies M = f_x \text{ and } N = f_y \implies M dx + N dy = f_x \ dx + f_y \ dy = df$$
Later, however, in a video, it was stated that the above was only true within simply connected regions. Does the above argument work in non-simply connected regions? Or is there simply something I'm missing in my argument?
After doing further research throughout the readings of 18.02 and 18.02sc, I've found that the above is indeed true in a non-simply connected region. The nuance I was missing was almost notational, in a sense - much like how $ \int_C \vec F \cdot d \vec r = \int_C M dx + N dy$ takes a dot product and carries it into a differential, $ \int_C \nabla f \cdot d \vec r = \int_C \ f_x \ dx + f_y \ dy$ appears to take $\nabla f$ and "split it apart."
As such, the difference between $\nabla f = f_x \ \vec i + f_y \ \vec j$ and $df = f_x \ dx + f_y \ dy$ is almost merely notational, as the two appear to be equivalent everywhere - the latter the differential analogue for the former.