If `f(x)=[2x]sin3pix` then prove that `f'(k^(+))=6kpi(-1)^(k)`, (where [.] denotes the greatest integer function and `k in N).`

Explanation

To prove that \( f'(k^{+}) = 6k\pi (-1)^{k} \) for the function \( f(x) = [2x] \sin(3\pi x) \), where \([.]\) denotes the greatest integer function and \( k \in \mathbb{N} \), we will follow these steps:

### Step 1: Define the derivative at \( k^{+} \)

The derivative \( f'(k^{+}) \) can be defined using the limit:

\[
f'(k^{+}) = \lim_{h \to 0^{+}} \frac{f(k+h) - f(k)}{h}
...