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Math Help - Deriviative of function

  1. #1
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    Deriviative of function

    If a force F is defined by F=k(e^(-s) -6e^(-2s)), where S is the distance between two objects, then prove (d^(2)F)/(d(S)^2)=F-18ke^(-25).


    I'm not sure what d^2 is supposed to mean.. does it mean take the derivative and square it?

    Well I did take the derivative.. did not square it yet. Don't know if this is right but gave it a shot:

    (d^(2)F)/(dS^2)=-e^-s -6e^(-2s)+k(-e^(-s) -6e^(-2s) (-2))
    =e^-s(-6e^-2s+k(12e^-2s-e^-s)
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  2. #2
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    Quote Originally Posted by skeske1234 View Post
    If a force F is defined by F=k(e^(-s) -6e^(-2s)), where S is the distance between two objects, then prove (d^(2)F)/(d(S)^2)=F-18ke^(-25).


    I'm not sure what d^2 is supposed to mean.. does it mean take the derivative and square it?
    No, it is the second derivative
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  3. #3
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    Quote Originally Posted by running-gag View Post
    No, it is the second derivative
    Ok i did that: found the 2nd derivative of F

    I did not end up with the answer the wanted me to prove............

    my answer:

    (12e^(-2s)-e^(-s))[1-24e^(-2s)+e^(-s)]

    What did I do wrong? please demonstrate, ty
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  4. #4
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    F=k(e^(-s) -6e^(-2s))

    dF/ds = k(-e^(-s)+12e^(-2s))

    dF/ds = k(e^(-s)-24e^(-2s)) = F - 18e^(-2s)
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  5. #5
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    Quote Originally Posted by running-gag View Post
    F=k(e^(-s) -6e^(-2s))

    dF/ds = k(-e^(-s)+12e^(-2s))

    dF/ds = k(e^(-s)-24e^(-2s)) = F - 18e^(-2s)
    for the first derivative, did you not use the product rule? does k = 0?
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  6. #6
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    You don't need to use the product rule since k is a constant
    The derivative of (kf), where k is a constant and f a function, is kf'
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  7. #7
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    Quote Originally Posted by running-gag View Post
    You don't need to use the product rule since k is a constant
    The derivative of (kf), where k is a constant and f a function, is kf'
    how do you know if k is a constant or not
    and
    for your last line:

    dF/ds = k(e^(-s)-24e^(-2s)) = F - 18e^(-2s)

    How did you get the last answer
    F-18e^(-2s)

    How did that pop up all of a sudden from your previous answer?....
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  8. #8
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    Quote Originally Posted by skeske1234 View Post
    If a force F is defined by F=k(e^(-s) -6e^(-2s)), where S is the distance between two objects, then prove (d^(2)F)/(d(S)^2)=F-18ke^(-25).
    F = k(e^{-s} - 6e^{-2s})

    \frac{dF}{ds} = k(-e^{-s} + 12e^{-2s})

    \frac{d^2F}{ds^2} = k(e^{-s} - 24e^{-2s})

    \frac{d^2F}{ds^2} = k(e^{-s} - 6e^{-2s} - 18e^{-2s})

    \frac{d^2F}{ds^2} = \textcolor{red}{k(e^{-s} - 6e^{-2s})} - 18ke^{-2s}

    \frac{d^2F}{ds^2} = \textcolor{red}{F} - 18ke^{-2s}
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