Expert: Alon Mandes - 11/11/2009

Question
QUESTION: Hi,

I need the derivative of the following cumlative distrubution function with respect to x.

f(x) = A*(1-e^-8Bx)/(1+Ce^-8Bx)

Where A,B and C are constants, and e is the Exponential Function.

Thanks,

Matt

ANSWER:         A*(1-e^-8Bx)         A-Ae^(-8Bx)
f(x)= ------------------ = ---------------
        1+Ce^-8Bx          1+Ce^(-8Bx)


       [ -8BAe^(-8Bx) ][ 1+Ce^(-8Bx) ] - [ -8CBe6(-8Bx) ][ A-Ae^(-8Bx) ]
f'(x)= -------------------------------------------------------------------
         [ 1+Ce^(-8Bx) ]²


       [ -8BAe^(-8Bx) ][ 1+Ce^(-8Bx) ] - [ -8CBe6(-8Bx) ][ A-Ae^(-8Bx) ]
f'(x)= -------------------------------------------------------------------
         [ 1+Ce^(-8Bx) ]²


You may proceed from here.

Alon.















---------- FOLLOW-UP ----------

QUESTION: Many thanks,

Although I think you mean [8BAe^(-8Bx)] rather than [-8BAe^(-8Bx)] in the first term of the Numerator of f'(x).

Thanks again

Answer
       A-Ae^(-8Bx)
f(x)= ------------------
       1+Ce^(-8Bx)  

IF y=[f/g] then y'=[f'g-fg']/g^2 . So,
      
       [ 8Be^(-8Bx) ][ 1+Ce^(-8Bx) ] - [ -8CBe^(-8Bx) ][ A-Ae^(-8Bx) ]
f'(x)= -------------------------------------------------------------------
         [ 1+Ce^(-8Bx) ]²