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Math Help - Trig Derivative Problem.

  1. #1
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    Trig Derivative Problem.

    Hello everyone, any assistance would be greatly appreciated.

    Let f(x) = 12xsinxcosx . Find f'(\frac{-3\pi}{2}).

    I attempt to differentiate the function using the product rule:
    I use the product rule for sinx cosx
    sinx(-sinx)+cosx(cosx)=-sin^2x+cos^2x

    And then 12x(-sin^2x+cos^2x)
    =12x(-1)+(-sin^2x+cos^2x)(12)
    =-12x-12sin^2x+12cos^2x

    Using the derivative of f(x), I try to evaluate f'(\frac{-3\pi}{2}).
    -12(\frac{-3\pi}{2})-12sin^2(\frac{-3\pi}{2})+12cos^2(\frac{-3\pi}{2})

    I end up getting 18\pi-12

    However, the book answer is just 18\pi.
    Can someone tell me where I went wrong? Thanks in advance.
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  2. #2
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    Re: Trig Derivative Problem.

    You're best off if you use the identity \displaystyle \begin{align*} \sin{(2x)} \equiv 2\sin{(x)}\cos{(x)} \end{align*}. Then \displaystyle \begin{align*} f(x) = 12x\sin{(x)}\cos{(x)} = 6x\sin{(2x)} \end{align*}. Then by the product rule we have

    \displaystyle \begin{align*} f(x) &= 6x\sin{(2x)} \\ f'(x) &= 6\left[ 2x\cos{(2x)} + \sin{(2x)} \right] \\ f'\left( -\frac{3\pi}{2} \right) &= 6\left[ 2\left( -\frac{3\pi}{2} \right) \cos{( -3\pi )} + \sin{(-3\pi)} \right] \\ &= 6\left[ -3\pi \left( -1 \right) + 0 \right] \\ &= 18\pi \end{align*}
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    Re: Trig Derivative Problem.

    Thank you very much for the quick response Prove It.
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