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Math Help - calculus of modulus function

  1. #1
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    calculus of modulus function

    Find the gradient of the curve f(x) = |sin(2x)| at x = \frac {2 \pi}{3}
    How do I do this? I think i need to find a general equation but i'm not sure how...

    thanks!
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by 301909 View Post
    Find the gradient of the curve f(x) = |sin(2x)| at x = \frac {2 \pi}{3}
    How do I do this? I think i need to find a general equation but i'm not sure how...

    thanks!
    We have that

    f(\varphi)=|sin(3\varphi)|

    so

    f'(\varphi)=sign(\sin(3\varphi))\cdot\cos(3\varphi  )\cdot{3}

    So f'\bigg(\frac{2\pi}{3}\bigg)=sign(0)\cdot{3}
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by 301909 View Post
    Find the gradient of the curve f(x) = |sin(2x)| at x = \frac {2 \pi}{3}
    How do I do this? I think i need to find a general equation but i'm not sure how...

    thanks!
    |\sin(2x)|=\begin{cases}\sin(2x) \mbox{ if } n\pi \le x \le \frac{(2n+1)\pi}{2}\\ -\sin(2x) \mbox{ if } \frac{(2n+1)\pi}{2} < x < (n+1)\pi\end{cases} \forall n \in \mathbb{Z}

    So now taking the derivative we get

    f'(x)=\begin{cases}2\cos(2x) \mbox{ if } n\pi \le x \le \frac{(2n+1)\pi}{2}\\ -2\cos(2x) \mbox{ if } \frac{(2n+1)\pi}{2} < x < (n+1)\pi\end{cases} \forall n \in \mathbb{Z}

    f' \left( \frac{2\pi}{3} \right)=-2\cos\left( \frac{4\pi}{3}\right)=1

    Edit: I guess I am running slow today
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