Calculate the derivative $f(x)=\lfloor x\rfloor(\sin(\pi x))^{2}$

Question

I have a problem with this task because answer which I have does not match the right answer and I don't know where is a mistake.

My try:
For $x\in \mathbb Z$ $f'_{+}(x)=f'_{-}(x)=0$ so $f'(x)$ exist
For $x \in (n,n+1), n \in \mathbb Z, f(x)=n(\sin(\pi x))^{2}$
So\begin{align}f'(x)&=1\cdot n^{0}(\sin(\pi x))^{2}+n((\sin(\pi x))^{2})'\\&=(\sin(\pi x))^{2}+n2\sin(\pi x)\cos(\pi x)\\&=(\sin(\pi x))^{2}+\sin(2\pi x)n\end{align}
In the answer is: $f'(x)=n \pi\sin(2\pi x)$ and then $f'(x)=\lfloor x\rfloor\pi\sin(2\pi x)$ because for $x \in \mathbb Z$ $f'(x)$ also exist.

Why?

Answer 1

Why $n^0$? If, in $(n,n+1)$, $f(x)=n\sin^2(\pi x)$, then, in that interval$$f'(x)=2n\pi\sin(\pi x)\cos(\pi x)=n\pi\sin(2\pi x).$$Don't forget that $n$ is a constant here.

Answer 2

The function to be differentiated is

$$f(x) = \lfloor x\rfloor \sin(\pi x)^2$$

By the product rule we have

$$f'(x) =a(x) + b(x)$$

where

$$a(x) =\lfloor x\rfloor \left(2 \pi \sin(\pi x) \cos(\pi x) \right)=\lfloor x\rfloor \left( \pi \sin(2\pi x)\right) $$

$$b(x) = \sin(\pi x)^2 \left( \frac{d}{dx} \lfloor x\rfloor \right)$$

Now it is not difficult to see that the derivative of the floor function produces a Dirac delta function at each integer point.

Hence

$$\left( \frac{d}{dx} \lfloor x\rfloor \right)= \sum_{k=-\infty}^\infty \delta(x-k)$$

But since

$g(x) \delta(x-k) = g(k)$ and $\sin(\pi k)=0$ we have $b=0$

and the derivative is

$$f'(x) =\lfloor x\rfloor \pi \sin(2\pi x) $$

This derivative reproduces $f(x)$ on integrating. For example

$$\int_{-\frac{1}{2}}^{\frac{3}{2}} f'(x) \, dx = 2$$

which agrees with

$$f(\frac{3}{2}) - f(-\frac{1}{2}) = 1 - (-1) = 2$$

Remark

As an answer to a comment I'd like to add that my approach also treats the more general case where

$$f(x) = \lfloor x\rfloor h(x)$$

and

$$f'(x) = \lfloor x\rfloor h'(x) + h(x) \sum_{k=-\infty}^\infty \delta(x-k)\\= \lfloor x\rfloor h'(x) + \sum_{k=-\infty}^\infty h(k)\delta(x-k)$$