# Derivative of floor function [ 1 / x ]

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 $$0 < x < 1$$?

• 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 Jan 10, 2020 at 19:47
• Oh yeah, fixed it Jan 10, 2020 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.