# Quotient group of functions with same derivative

Let $H := \{ f: f \text{ differentiable}\}$ and the equivalence relation $\sim(f,g) := f' = g' \text{everywhere}$ with $f,g \in H$. I'm wondering about the nature of $H / \sim$.

Intuitively I would say all same functions with different additive constants at the end will be identified, is this correct? Furthermore if we let $f' = g'\text{ a.e.}$ would our classes change by a lot, since we have already required differentiability? And what about $f'' = g''$ and so on?

Let $U\subseteq \mathbb{R}$ be open and connected and $f,g : U \rightarrow \mathbb{R}$ differentiable functions such that $f'=g'$. Then we have by the fundamental theorem of calculus ( $f'-g'=0$ is continuous)
$$(f-g)(x) - (f-g)(x_0) = \int_{x_0}^x (f-g)'(y)dy = 0$$
for some $x_0\in U$. More generally, we have everywhere differentiable function whose derivative is 0 almost everywhere is a constant. Which is telling us that functions that get identified under your equivalence relation are precisely those, which on every connected component differ by some constant.
For higher derivatives, say $n$ times differentiable, you can play the same game (just integrate some more times) and get that exactly those functions get identified which on every connected component differ by a polynomial of degree less or equal to $n-1$.