Find the $n$th derivative of $f(x)=x\sin(x)\cos(2x)$ If it helps, it ask the value for $n=100$ and $x=\pi/2$.
I can't do it by induction because it has too many factors and trying to use an equality for $\cos(2x)$ didn't helped.
I don't see the relation in the derivatives.
 A: First note that
$$
  \left(\frac{d}{dx}\right)^n(xg(x))=x\left(\frac{d}{dx}\right)^ng(x)
    +n\left(\frac{d}{dx}\right)^{n-1}g(x)
$$
(you can check this by induction). Secondly your function can be written
$$
  f(x)=x\times\frac12\left(\sin(3x)-\sin(x)\right).
$$
Thus
$$
  \left(\frac{d}{dx}\right)^{100}f(x)
    =\frac x2\left(3^{100}\sin(3x)-\sin(x)\right)
      +\frac{100}2\left(-3^{99}\cos(3x)+\cos(x)\right).
$$
Now you can substitute $x=\frac{\pi}2$.
A: Observe
\begin{align}
x\sin x \cos 2x =&\  x\frac{e^{ix}-e^{-ix}}{2i} \frac{e^{2ix}+e^{-2ix}}{2} = \frac{x}{4i}\left(e^{3ix}-e^{ix}+e^{-ix}-e^{-3ix} \right)\\
=&\ \frac{1}{2}x\sin 3x-\frac{1}{2}x\sin x \\
=&\ \frac{1}{2}\left(x-\frac{\pi}{2}\right)\sin 3x -\frac{1}{2}\left(x-\frac{\pi}{2}\right)\sin x + \frac{\pi}{4}\sin 3x -\frac{\pi}{4}\sin x\\
=&\ -\frac{1}{2}\left(x-\frac{\pi}{2}\right)\cos\left(3 \left(x-\frac{\pi}{2} \right) \right)- \frac{1}{2}\left(x-\frac{\pi}{2}\right)\cos \left(x-\frac{\pi}{2}\right) \\
&- \frac{\pi}{4}\cos\left(3 \left(x-\frac{\pi}{2} \right) \right)-\frac{\pi}{4}\cos\left(x-\frac{\pi}{2}\right). 
\end{align}
Since
\begin{align}
\cos\left(3 \left(x-\frac{\pi}{2} \right) \right)=\sum^\infty_{k=0} (-1)^k\frac{3^{2k}(x-\pi/2)^{2k}}{(2k)!}
\end{align}
and
\begin{align}
\cos \left(x-\frac{\pi}{2}  \right)=\sum^\infty_{k=0} (-1)^k\frac{(x-\pi/2)^{2k}}{(2k)!}
\end{align}
then it follows the 
\begin{align}
f^{(100)}(\pi/2) = -\frac{\pi}{4}100! \left(\frac{1}{100!}+\frac{3^{100}}{100!} \right) = -\frac{\pi}{4}(3^{100}+1). 
\end{align}
