What is the derivative of the floor function below w.r.t $x$ from first principle.

$\left[ \frac{1}{x} \right]$ , where $x \mapsto [x]$ represents floor function

I think the derivative only exist for values of $1/x \notin \Bbb N$ but does the derivative exit for $0 < x < 1$?

  • 1
    $\begingroup$ The derivative is $0$ for all $x$ such that $1/x$ is defined and $1/x \notin \Bbb N$. For $1/x \in \Bbb N$ the limit only exists from the right, and you might say that the function has a "right derivative". See here $\endgroup$ – Norse Jan 10 '20 at 19:47
  • $\begingroup$ Oh yeah, fixed it $\endgroup$ – Norse Jan 10 '20 at 19:50

The derivative of the floor function is always $0$ except at the points where $\frac 1n{\in I}$ where the graph is discontinuous.


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