Find formula of sum $\sin (nx)$ I wonder if there is a way to calculate the
$$S_n=\sin x + \sin 2x + … + \sin nx$$ 
but using only derivatives ? 

Using telescopic sums:
$$ \sin(mx)\sin(x/2) = \frac{1}{2}\left(\cos\left((m-1/2)x\right)-\cos\left((m+1/2)x\right)\right)$$
Hence:
$$ S_n \sin\frac{x}{2} = \frac{1}{2}\left(\cos\frac{x}{2}-\cos\left(\left(n+\frac{1}{2}\right)x\right)\right)=\sin\frac{nx}{2}\cdot\sin\frac{(n+1)x}{2}.$$
