Derivative of a smooth function as an equivalence class In my differential manifolds class, the derivative of a function $f \in C^\infty (M)$, where $M$ is a manifold, at $p \in M$ was defined as the image of the linear map
$$(df)_p := C^\infty (M) \mapsto C^\infty (M)/Z_p$$
where $Z_p$ is the set of all smooth functions that have zero derivative at $p$. In other words, the derivative is an equivalence class of $C^\infty (M)$ defined by the relation that two functions are related if their derivative vanishes at $p$, the derivative of a function at a point is an equivalence class.
I'm struggling to wrap my head around this. How can a derivative become an equivalence class? Does this hold in elementary calculus too?
 A: The practice of defining mathematical objects as quotients, i.e. sets of equivalence classes of some other, more elementary objects, is widespread in mathematics. Here are a few examples: homology (where two chains are considered equivalent if they share a boundary), projective space (where two vectors are considered equivalent if one is a scalar multiple of the other) and rational numbers (where two pairs of suitable integers $(a, b)$ and $(c, d)$ are considered equivalent if $ad = bc$).
One way to help wrap one's mind around this is to distinguish between and adopt two perspectives. In one view, you consider an object's "internal structure", i.e. how it is built from other, more elementary objects. In the other view, you consider the object's "external properties", i.e. what relationships it enters with other abstract objects.
You generally use the first perspective to derive basic properties and then you begin treating the object as a "black box", forgetting how it has been built and just using its known properties to derive more properties. (An extreme form of the latter perspective occurs when you postulate a set of basic properties as axioms. This allows you to build a theory that applies to any construction that satisfies them.)

Responding more directly to your question: I believe the definition you have in mind is not in fact a definition of a derivative, but one of the definitions of the contangent space. In any case, in this setting, the first perspective tells you how the contangent space and the differential 1-forms that make it up are constructed and allows you to establish that they have some basic properties (e.g. that they satisfy the Leibniz product rule). Soon after that you generally stop thinking much about their internal structure and adopt the second perspective.

Regarding the relationship to elementary calculus: differential geometry intentionally raises the level of abstraction to increase the power of the theory beyond what you can achieve in elementary calculus. Many concepts familiar from calculus become special cases for the concepts in differential geometry. In particular, in calculus you can exploit the linear structure of $\mathbb{R}^n$ to pretend that tangent vectors live within the manifold under study. This allows you to maintain certain comfortable illusions, e.g. that a derivative of a smooth function $\mathbb{R} \to \mathbb{R}$ is also a function $\mathbb{R} \to \mathbb{R}$.
A: Adam Zalcman's answer is already really great so I don't have much to add, except for an explanation of why the differential has this form; which I found in Introduction to Smooth Manifolds by John Lee (chapter 11) and I found quite nice:


A: Maybe you are over-interpreting it a bit. It just says that if $f=g+h$ with $\frac{\partial h}{\partial {x}^i}=0$ for any chart, you have $f\in [g]$. In other words, $[f]=[g]$ iff $\frac{\partial f}{\partial {x}^i}=\frac{\partial g}{\partial {x}^i}$. It's just a way to express the idea of $df=\frac{\partial f}{\partial {x}^i} dx^i$ without coordinate frames.
