Derivative of $\lceil 1/x \rceil$ I'm looking for the derivative $\frac{d}{dx}\lceil 1/x \rceil$.
I would like to find a real number $1<x \le y$ satisfying the minimum of $\left\lceil \frac{y}{x} \right\rceil x$, when $y$ is a fixed value $>0$.
Is the ceiling function a problem ?

$\large x > 0:$

\begin{align}
\left\lceil x\right\rceil
&=
\Theta\left(x\right)\Theta\left(1 - x\right)
+
2\Theta\left(x - 1\right)\Theta\left(2 - x\right)
+
3\Theta\left(x - 2\right)\Theta\left(3 - x\right) + \cdots
\\[3mm]&=
\sum_{n = 0}^{\infty}\left(n + 1\right)\Theta\left(x - n\right)
\Theta\left(n + 1 - x\right)
\end{align}


\begin{align}
{{\rm d}\left\lceil x\right\rceil \over {\rm d}x}
&=
\sum_{n = 0}^{\infty}\left(n + 1\right)\left[%
\delta\left(x - n\right)
\Theta\left(n + 1 - x\right)
-
\Theta\left(x - n\right)
\delta\left(n + 1 - x\right)
\right]
\\[3mm]&=
\sum_{n = 0}^{\infty}\left(n + 1\right)\left[%
\delta\left(x - n\right)
\Theta\left(1\right)
-
\Theta\left(1\right)
\delta\left(n + 1 - x\right)
\right]
\\[3mm]&=
\sum_{n = 0}^{\infty}\left(n + 1\right)
\delta\left(x - n\right)
-
\sum_{n = 1}^{\infty}
n\,\delta\left(n - x\right)
=
\sum_{n = 0}^{\infty}\delta\left(x - n\right)
\end{align}


\begin{align}
{{\rm d}\left\lceil x\right\rceil \over {\rm d}x}
&=
\sum_{n = 0}^{\infty}\delta\left(x - n\right)\,,
\qquad
\left\lceil x \right\rceil = 1\ \mbox{when}\ 0 < x < 1
\end{align}


\begin{align}
{{\rm d}\left\lceil 1/x\right\rceil \over {\rm d}x}
&=
\sum_{n = 0}^{\infty}\delta\left({1 \over x} - n\right)\,
\left(-\,{1 \over x^{2}}\right)
=
\sum_{n = 1}^{\infty}
{\delta\left(x - 1/n\right) \over \left\vert -1/x^{2}\right\vert}\,
\left(-\,{1 \over x^{2}}\right)
\end{align}


\begin{align}
\color{#ff0000}{\large{{\rm d}\left\lceil 1/x\right\rceil \over {\rm d}x}}
&\color{#000000}{\large\ =\ }
\color{#ff0000}{\large -\sum_{n = 1}^{\infty}\delta\left(x - {1 \over n}\right)}
\end{align}

