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Need help with Logistical Derivative Problem

  1. May 1, 2007 #1
    1. The problem statement, all variables and given/known data

    -5.978(1+5.71e^-.054x)/1+5.71e^-.054x

    logistical regression values from ti 83

    a=5.71
    b=.054
    c=5.978



    2. Relevant equations
    general logistical regresion model g(x) c/1+ae^-bx

    3. The attempt at a solution

    So far I have applied the quotient rule to find the derivative of the general logistical model and come out to

    (1+ae^-bx)-c(1+ae^-bx)/(1+ae^-bx)^2

    I took derivate and factored a little and came to this

    -c(1+ae^-bx)/1+ae^-bx

    At this point I just factored in the logistical information from my ti 83 which is

    a=5.71
    b=.054
    c= 5.978


    I really need help factoring the numerator of this problem, and from there finding the 2nd derivative. I know at some point I must take natural log to both side of the equation to find the point of inflection. Please help me with this!
     
  2. jcsd
  3. May 1, 2007 #2

    cristo

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    What's the actual question here?

    From what I can gather you've tried to differentiate [tex]\frac{c}{1+ae^{-bx}}[/tex]. IF that's what you've done, then you've done it incorrectly. You dont really need the quotient rule for this; write the expression as [tex]c(1+ae^{-bx})^{-1}[/tex] and differentiate via the chain rule to give [tex]-c(1+ae^{-bx})^{-2}\cdot -abe^{-bx}=abce^{-bx}(1+ae^{-bx})^{-2}[/tex]

    As for where to go from here, I can't advise, as I don't know what you're trying to do!
     
    Last edited: May 1, 2007
  4. May 1, 2007 #3
    Thanks alot cristo. Basicaly I need help factoring this

    -5.978(1+5.71e^-.054x)/1+5.71e^-.054x

    so I can find the second derivative
     
  5. May 1, 2007 #4

    Integral

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    Please do not double (or triple) post. Read Cristo's post. You have not got the 1st derivitive correct so the expression you are trying to factor is meaningless.
     
  6. May 1, 2007 #5
    Yes, I must have read it incorectly. Pluging in I now get

    5.71(.054)5.978e^-.054x(1+5.71e^-.054x)^-2

    Can someone help me out with this. Sorry for being so progressive.
     
  7. May 1, 2007 #6

    cristo

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    If I were you, I'd leave the constants as letters and then plug them in later (they're not going to change through differentiation). So, you want to differentiate [tex]\frac{abce^{-bx}}{(1+ae^{-bx})^2}[/tex]. To do this you will need the quotient rule. Recall that it says the derivative of a quotient is [tex]\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}[/tex], where the primes denote differetiation wrt x.

    In your question, take [itex] u=abce^{-bx}, \text{and} \;\;\;v=(1+ae^{-bx})^2[/itex]. Have a go and post your efforts and we'll point you in the right direction.
     
  8. May 1, 2007 #7
    I don't recognize some of the characters/symbols u are using. I don't think I am there yet. but pluging in my equations i get 1.84e^-.054x/(1+ae^-.054x)^2 . I think the e symbol is what is throwing me off a little.
     
  9. May 1, 2007 #8

    cristo

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    e is the exponential function-- you used it in your original post, and I would take that as assumed knowledge for a calculus course.

    Recall: [tex]\frac{d}{dx}e^{Ax}=Ae^{Ax}[/tex] where A is any constant.

    I don't know what you've done here. Show some working that I can follow!
     
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