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Math Help - Need some help finding the 2nd deriv of this problem.

  1. #1
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    Need some help finding the 2nd deriv of this problem.

    F(theta)= (cos(theta pi))^ -2 find 2nd derivative of F(2). I have tried it different ways. I have worked it from the original to the 1st then 2nd derivative of theta. At the end I plugged 2 in and that was no good. I tried taking derivative of 2 from the 1st to the second and that hasnt worked either. Can some please explain where I am wrong. If i replace theta with 2 then would it not be (cos(2pi))^ -2 and wouldn't that be the same as 1/((cos(2 pi))^2 and then use the quotient rule for the 1st derivative and go from there? Who knows my brain is fried and help would be awesome.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Need some help finding the 2nd deriv of this problem.

    We are given:

    F(\theta)=(\cos(\pi\theta))^{-2}

    and we are asked to find:

    F''(2)

    To find the first derivative, we may use the power and chain rules:

    F'(\theta)=-2(\cos(\pi\theta))^{-3}(-\pi\sin(\pi\theta))=2\pi\sin(\pi\theta)(\cos(\pi \theta))^{-3}

    To find the second derivative, we may use the product, power and chain rules:

    F''(\theta)=2\pi\left(\sin(\pi\theta)\left(-3(\cos(\pi\theta))^{-4}(-\pi\sin(\pi\theta)) \right)+\pi\\cos(\pi\theta)(\cos(\pi\theta))^{-3} \right)

    F''(\theta)=2\pi^2(\cos(\pi\theta))^{-4}\left(3\sin^2(\pi\theta)+\cos^2(\pi\theta) \right)

    F''(\theta)=2\pi^2\sec^4(\pi\theta)\left(2-\cos(2\pi\theta) \right)

    Hence:

    F''(2)=2\pi^2
    Last edited by MarkFL; September 28th 2012 at 10:56 PM. Reason: to fix LaTeX issue
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  3. #3
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    Re: Need some help finding the 2nd deriv of this problem.

    Thank you very much. Im still a little bit confused how did you factor out the pi in the first deriv. -pi sin (theta pi)
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Need some help finding the 2nd deriv of this problem.

    It stems from the application of the chain rule.

    \frac{d}{d\theta}\left((cos(\pi\theta)^{-2} \right)=-2(cos(\pi\theta)^{-3}\frac{d}{d\theta}\left(\cos(\pi\theta) \right)

    \frac{d}{d\theta}\left((cos(\pi\theta)^{-2} \right)=-2(cos(\pi\theta)^{-3}(-\sin(\pi\theta))\frac{d}{d\theta}(\pi\theta)

    \frac{d}{d\theta}\left((cos(\pi\theta)^{-2} \right)=-2(cos(\pi\theta)^{-3}(-\sin(\pi\theta))(\pi)

    \frac{d}{d\theta}\left((cos(\pi\theta)^{-2} \right)=2\pi\sin(\pi\theta)(cos(\pi\theta)^{-3}
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