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Math Help - Second Derivative Problem

  1. #1
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    Second Derivative Problem

    given that h(x) = \frac{1}{(3x-5)^2^'}.Find h''(1)
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  2. #2
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    Have you tried the chain rule? what did you get?
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  3. #3
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    i have no idea sir.I confused with the 2 prime sign.
    Last edited by faraday; May 21st 2011 at 10:26 AM.
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  4. #4
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    "double prime" just means differentiate twice, ie obtain the derivative, then differentiate the derivative.
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  5. #5
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    i mean the prime at the fraction.
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  6. #6
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    i assumed that was a typo! It has no standard meaning that i'm aware of.

    If it has a meaning that is particular to your course/teacher then the answer is probably in your course notes or textbook.
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  7. #7
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    ok..when i using chain rule.I get.
    h(x) = (3x-5)^{-2}

    \frac{dy}{dx} = 3(-2) (3x-5) ^ {-2-1} \times \frac{d}{dx} (3x-5)

    =-6(3x-5)^{-3} \times 3

    = \frac{-18}{(3x-5)^3}

    is that correct?then what should i do?
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  8. #8
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    that isn't correct. You pulled out a factor of "3" on your second line that isn't needed. h'(x) should be:

    h'(x) = \dfrac{dy}{dx} = (-2)(3x-5)^{-2-1) \times \dfrac{d}{dx}(3x-5)

    h'(x) = \dfrac{dy}{dx} = (-2)(3x-5)^{-3) \times 3

    h'(x) = \dfrac{dy}{dx} = -6(3x-5)^{-3)

    h'(x) = \dfrac{dy}{dx} = \frac{-6}{(3x-5)^3}

    Differentiate again to get h''(x). Then evaluate your answer at x=1 to find h''(1).
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