TITLE: Geometric Interpretation of Weak Derivative
QUESTION [7 upvotes]: As we know, classic derivative $f'(x)$ of a function $f(x)$ can be interpreted as the rate of change of function $f$ in each point $x.$
How about weak derivative? Since it is defined through integral and therefore not relevant in sets of zero measure, what does it mean for a function to have weak derivative? What can the weak derivative of a function explains about its function?

REPLY [5 votes]: If $g$ is the weak derivative of $f$, then the Fundamental Theorem of Calculus holds: 
$$f(b)-f(a) = \int_a^b g(x)\,dx \tag{1}$$
Thus, even though $g$ may not succeed in expressing the infinitesimal rate of change of $f$ at every point, it captures the rate of change of $f$ on every non-infinitesimal scale. 
For the purpose of estimating $f$, property $(1)$ is about as good as having classical derivative. For example, if $g$ happens to be square-integrable, we can write 
$$|f(b)-f(a)| \le \int_a^b |g(x)|\,dx \le\sqrt{b-a} \sqrt{\int_a^b |g(x)|^2\,dx}$$
and conclude that $f$ is Hölder continuous with exponent $1/2$. 
For some other purposes, like locating the maximum or minimum of $f$, the weak derivative is not the right tool. For that one would use another generalized concept of derivative (Subderivative).