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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.

Sloan
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Arji
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  • It would seem like $(-1)^{\lfloor n/2\rfloor}$, where $n$ is the order of the derivative you are trying to take, will take care of that particular bit. – Nick Peterson Sep 14 '16 at 16:17
  • See http://math.stackexchange.com/questions/1752455/100-th-derivative-of-the-function-fx-ex-cosx/1752462#1752462 – lab bhattacharjee Sep 14 '16 at 16:17

6 Answers6

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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}

celtschk
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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.

Michael Hardy
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John
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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 $$f^{(2n)}=(-1)^n2^{2n}\sin(2x).$$

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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

$$1,2i,-4,-8i,16,32i,-64,-128i,\cdots$$

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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}

Michael Hardy
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$$y=\sin 2x$$ $$y'=2\cos 2x$$ $$y''=-4\sin 2x \Rightarrow y''=-4y$$ $$y'''=-4y'$$ $$y''''=-4y''\Rightarrow y''''=-4(-4y)=(-4)^2y$$ conclude that for even derivative $$y^{2n}=(-4)^n\sin 2x\tag1$$ to get the odd derivatives, we can derive $eq.1$ one time $$y^{2n+1}=2(-4)^n\cos 2x$$

E.H.E
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