Thread: The n-th derivative of x*cos(2x)

1. The n-th derivative of x*cos(2x)

Hey, I have to find the $\displaystyle f^{(n)}(x)$ of function $\displaystyle f(x) = x \cdot \cos(2x)$. I know I need to find a Leibnitz formula for derivative of product, but how?

Also, I have to solve this formula explicitly, so, $\displaystyle f^{(n)}(x) = ...$.

2. Re: The n-th derivative of x*cos(2x)

start with a generalized product rule by letting $u = \cos(2x)$

$f(x) = x \cdot u$

$f'(x) = \color{red}{x \cdot u' + u}$

$f''(x) = x \cdot u'' + u' + u' = \color{red}{x \cdot u'' + 2u'}$

$f'''(x) = x \cdot u''' + u'' + 2u'' = \color{red}{x \cdot u''' + 3u''}$

$f^{(4)}(x) = x \cdot u^{(4)} + u''' + 3u''' = \color{red}{x \cdot u^{(4)} + 4u'''}$

... $f^{(n)}(x) = \color{red}{x \cdot u^{(n)} + nu^{(n-1)}}$

now for the fun part,

$u = \cos(2x)$

$u' = -2\sin(2x)$

$u'' = -4\cos(2x)$

$u''' = 8\sin(2x)$

$u^{(4)} = 16\cos(2x)$

...

note that both $\pm \cos(2x)$ and $\pm \sin(2x)$ can be written in the form $\cos(2x+b)$ where $b$ is a multiple of $\dfrac{\pi}{2}$

... $u^{(n)} = 2^n \cos(2x+b) = 2^n\bigg[\cos(2x)\cos{b} - \sin(2x)\sin{b}\bigg]$

$n=1 \implies b = \dfrac{\pi}{2}$

$n=2 \implies b = \dfrac{2\pi}{2}$

$n=3 \implies b = \dfrac{3\pi}{2}$

$n=4 \implies b = \dfrac{4\pi}{2}$

and the cycle continues

... $u^{(n)} = 2^n \cos\left(2x + \dfrac{n\pi}{2}\right)$