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

  1. #1
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    Exclamation 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, $\displaystyle f^{(n)}(x) = ...$.

    Please help
    Last edited by lebdim; Jun 30th 2019 at 03:54 AM.
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  2. #2
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    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


    ... $u^{(n)} = 2^n \cos\left(2x + \dfrac{n\pi}{2}\right)$
    Thanks from topsquark and lebdim
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