Find $n$-th derivative of $\sin(2x)$ I'm looking for the $n$-th derivative of $f(x) = \sin(2x)$. I build the first derivatives and tried to find a pattern and I did, but I did not find a function for that pattern. Here are the first derivatives:
\begin{align}
f'(x) & = \phantom{-}2 \cos(2x) \\
f''(x) & = -4 \sin(2x) \\
f'''(x) & = -8 \cos(2x) \\
& \,\,\,\vdots
\end{align}
The inner function $2x$ stays the same. I do not know how the coefficient can change from positive to negative but only every two derivatives. 
 A: If you want a single formula parametrized with $n$, try
$$f^{(n)}(x) = 2^n\sin\left(2x+n\frac{\pi}{2}\right)$$
This works because
\begin{aligned}
\sin \left(x+\frac{\pi}{2}\right)  &= \phantom{-}\cos x =\sin' x\\
\cos \left(x+\frac{\pi}{2}\right)  &= -\sin x =\cos' x
\end{aligned}
A: There are a few ways to capture the sign change.  Here's a hint.
Look at the behavior of $(-1)^{\lfloor n/2\rfloor}$
As for switching between $\sin(2x)$ and $\cos(2x)$, find something similar that alternates between $1$ and $0$.  (Actually, $\sin^2(n\pi / 2)$ kinda looks promising.)
Or, you can just enumerate four cases for each possibility of $n \bmod 4$ and be done more quickly.
A: Handle the even and odd order derivatives apart.
$$
f'(x)  = \phantom{-}2 \cos(2x) \\
f'''(x)  = -8 \cos(2x) \\
f'''''(x)  = 32 \cos(2x)\\
\cdots
$$
and $$f^{(2n+1)}=(-1)^n2^{2n+1}\cos(2x).$$
Then
$$f(x)=\sin(2x)\\
f''(x)  = -4 \sin(2x) \\
f''''(x)  = 16 \sin(2x) \\
\cdots
$$
and $$f^{(2n)}=(-1)^n2^{2n}\sin(2x).$$
A: Using Euler's formula,
$$
e^{i2x}=\cos(2x)+i\sin(2x),\\
\left(e^{i2x}\right)'=-2\sin(2x)+2i\cos(2x)=2ie^{i2x},\\
\left(e^{i2x}\right)''=-4\cos(2x)-4i\sin(2x)=(2i)^2e^{i2x},\\
\cdots
$$
and obviously
$$\left(e^{i2x}\right)^{(n)}=(2i)^ne^{i2x}$$
and
$$\left(\sin(2x)\right)^{(n)}=\Im((2i)^ne^{i2x}).$$
Notice the values of the factor, with alternating signs
$$1,2i,-4,-8i,16,32i,-64,-128i,\cdots$$
A: Once you have $\sin'=\cos$ and $\cos'=-\sin$ then you've got alternation of signs but only every two steps, because next you get $(-\sin)'=-\cos$ and then $(-\cos)'=-\sin$.
\begin{align}
\sin' & = \cos \\[8pt]
\cos' & = -\sin \\[8pt]
(-\sin)' & = -\cos \\[8pt]
(-\cos)' & = \sin \\[8pt]
\sin' & = \cdots & \text{Here you've returned to where you started.}
\end{align}
A: $$y=\sin 2x$$
$$y'=2\cos 2x$$
$$y''=-4\sin 2x \Rightarrow y''=-4y$$
$$y'''=-4y'$$
$$y''''=-4y''\Rightarrow y''''=-4(-4y)=(-4)^2y$$
conclude that for even derivative
$$y^{2n}=(-4)^n\sin 2x\tag1$$
to get the odd derivatives, we can derive $eq.1$ one time
$$y^{2n+1}=2(-4)^n\cos 2x$$
