I'm looking to calculate the derivative of the Fejer kernel: $$ F_n(x) = \frac1{n+1}\left(\frac{\sin\left(\frac{n+1}{2}x\right)}{\sin\left(\frac x2\right)}\right)^2 $$ However, I'm running into some real difficulty with the calculation. Supposedly I should get $F_n' = 2n\sin(nx)F_{n-1}(x)$ (or something like this), but I'm not getting anywhere near this result and I'm not sure why.
Just via rote calculation, using $\left(\frac uv\right)' = \frac{u'v - uv'}{v^2}$, we calculate $$ \begin{align} F_n' & = \frac1{n+1}\frac{\sin((n+1)x/2)}{\sin^3(x/2)}\left[(n+1)\cos((n+1)x/2)\sin(x/2) - \sin((n+1)x/2)\cos(x/2)\right] \\ & = \frac1{n+1}\frac{\sin((n+1)x/2)}{\sin^3(x/2)}\left[n\cos((n+1)x/2)\sin(x/2)-\sin(nx/2)\right] \end{align} $$ where the simplification on the second line is simply $\sin(x - y) = \sin x\cos y - \cos x\sin y$.
Frankly I have no clue what to do from here. If the formula $F_n' = 2(n+1)\sin((n+1)x)F_{n-1}(x)$ is true, then supposedly $$ \frac{2(n+1)}{n}\frac{\sin((n+1)x)\sin^2(nx/2)}{\sin^2(x/2)} = \frac1{n+1}\frac{\sin((n+1)x/2)}{\sin^3(x/2)}[n\cos((n+1)x/2)\sin(x/2)-\sin(nx/2)] $$ but I see absolutely no reason why this should be the case. In fact, plugging both sides into Desmos for some $n$ gives completely different graphs, validating my suspicions.
Is there actually a nice formula for the derivative of the Fejer kernel? If so, what is it? I can't see how to simplify my result further.