Solutions

Notice that for $x\notin\mathbb{Z}$, we have

$$\begin{align*} f'(x)&=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \\ &=\lim_{h\to 0}\frac{\lfloor x+h\rfloor-\lfloor x\rfloor}{h} \\ &=\frac{x-x}{h} \\ &=0. \end{align*}$$

For $x\in\mathbb{Z}$, we have

$$\begin{align*} \lim_{h\to 0^-}\frac{f(x+h)-f(x)}{h} &=\lim_{h\to 0^-}\frac{\lfloor x+h\rfloor-\lfloor x\rfloor}{h} \\ &=\lim_{h\to 0^-}\frac{x-1-x}{h} \\ &=\lim_{h\to 0^-}\frac{-1}{h}. \end{align*}$$

Hence, the left hand derivative does not exist for $x\in\mathbb{Z}$. Consequently, the derivative does not exist for $x\in\mathbb{Z}$.