Question

use the first principle definition to find the derivative of f'(a), where [x] is the greatest integer less than or equal to x.

Answer 1

1 STEP: We give a graph of this function $y=\lfloor x\rfloor$ so that it is easier to consider the derivative by definition

(More information: https://en.wikipedia.org/wiki/Floor_and_ceiling_functions)

2 STEP: There are 2 cases to consider.

1 case: $x\in(n;n+1), x\notin\mathbb{Z}\longrightarrow\lfloor x\rfloor=n$ , so

$$\frac{d}{dx}\lfloor x\rfloor=\lim\limits_{h\to0}\frac{\lfloor x+h\rfloor-\lfloor x\rfloor}{h}=\lim\limits_{h\to0}\frac{n-n}{h}=\lim\limits_{h\to0}\frac{0}{h}=0$$

Conclusion,

$$\boxed{\frac{d}{dx}\lfloor x\rfloor=0,\quad\text{for}\quad x\notin\mathbb{Z}}$$

2 case: $x\in\mathbb{Z}\longrightarrow x=n$ , so

$$\lim\limits_{h\to0^-}\frac{\lfloor x+h\rfloor-\lfloor x\rfloor}{h}=\lim\limits_{h\to0^-}\frac{\left(n-1\right)-n}{h}=\lim\limits_{h\to0^-}\frac{-1}{h}=+\infty\\[0.3cm] \lim\limits_{h\to0^+}\frac{\lfloor x+h\rfloor-\lfloor x\rfloor}{h}=\lim\limits_{h\to0^+}\frac{n-n}{h}=\lim\limits_{h\to0^-}\frac{0}{h}=0\\[0.3cm] \lim\limits_{h\to0^-}\frac{\lfloor x+h\rfloor-\lfloor x\rfloor}{h}=+\infty\neq0=\lim\limits_{h\to0^+}\frac{\lfloor x+h\rfloor-\lfloor x\rfloor}{h}\\[0.3cm] \nexists\lim\limits_{h\to0}\frac{\lfloor x+h\rfloor-\lfloor x\rfloor}{h}\longrightarrow\\[0.3cm]\boxed{\nexists\frac{d}{dx}\lfloor x\rfloor,\quad\text{for}\quad x\in\mathbb{Z}}$$

General conclusion,

$$\frac{d}{dx}\lfloor x\rfloor=\left[\begin{array}{l} 0,\quad\text{for}\quad x\notin\mathbb{Z}\\ DNE,\quad\text{for}\quad x\in\mathbb{Z} \end{array}\right.$$