Answer
$f$ is not differentiable for integers, but $f$ is differentiable for non-integers.
$f'(x) = 0~~$ if $x$ is a non-integer.
We can see a sketch of $f'(x)$ below.
Work Step by Step
Suppose $b$ is not an integer.
Let $\lfloor b \rfloor = c$
Then:
$f'(b) = \lim\limits_{h \to 0}\frac{f(b+h)-f(b)}{h}$
$f'(b) = \lim\limits_{h \to 0}\frac{\lfloor b+h \rfloor-\lfloor b \rfloor}{h}$
$f'(b) = \frac{c-c}{h}$
$f'(b) = \frac{0}{h}$
$f'(b) = 0$
$f'(x)$ exists for non-integers so $f$ is differentiable for non-integers.
Suppose $b$ is an integer.
$f(b) = \lfloor b \rfloor = b$
$\lim\limits_{x \to b^-}f(x) = \lim\limits_{x \to b^-}\lfloor x \rfloor = b-1$
$\lim\limits_{x \to b^-}f(x) \neq f(b),~~$ so $f$ is not continuous at $b$
Since $f$ is not continuous at $b$, $f$ is not differentiable at $b$.
Therefore, $f$ is not differentiable for integers.
$f$ is not differentiable for integers, but $f$ is differentiable for non-integers.
$f'(x) = 0~~$ if $x$ is a non-integer.
We can see a sketch of $f'(x)$ below.
