TITLE: Can derivatives be defined as anti-integrals?
QUESTION [32 upvotes]: I see integrals defined as anti-derivatives but for some reason I haven't come across the reverse. Both seem equally implied by the fundamental theorem of calculus.
This emerged as a sticking point in this question.

REPLY [5 votes]: Weak derivatives.
This is essentially the way one defines a weak derivative. If a function is not differentiable in the traditional sense, but it is integrable, then one may define a weaker notion of derivative through duality: the derivative of $f$ is the function $f'$ such that
$$
\int f' u=-\int f u'
$$
for all smooth functions $u$. One can prove that the function $f'$ is in fact $L^p$-unique. If $f$ is differentiable in the standard sense, then it is also differentiable in the weak sense, and both derivatives agree.
For example, the Dirichlet function is nowhere continuous, let alone differentiable. But its weak derivative exists, and is in fact the zero function. Indeed,
$$
0=\int 1_{\mathbb Q} u'=-\int 1'_{\mathbb Q} u
$$
implies that $1'_{\mathbb Q}=0$ almost everywhere.