Need help with Logistical Derivative Problem

1. May 1, 2007

goatrance

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. May 1, 2007

cristo

Staff Emeritus
What's the actual question here?

From what I can gather you've tried to differentiate $$\frac{c}{1+ae^{-bx}}$$. 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 $$c(1+ae^{-bx})^{-1}$$ and differentiate via the chain rule to give $$-c(1+ae^{-bx})^{-2}\cdot -abe^{-bx}=abce^{-bx}(1+ae^{-bx})^{-2}$$

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
3. May 1, 2007

goatrance

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

4. May 1, 2007

Integral

Staff Emeritus
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.

5. May 1, 2007

goatrance

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.

6. May 1, 2007

cristo

Staff Emeritus
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 $$\frac{abce^{-bx}}{(1+ae^{-bx})^2}$$. To do this you will need the quotient rule. Recall that it says the derivative of a quotient is $$\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}$$, where the primes denote differetiation wrt x.

In your question, take $u=abce^{-bx}, \text{and} \;\;\;v=(1+ae^{-bx})^2$. Have a go and post your efforts and we'll point you in the right direction.

7. May 1, 2007

goatrance

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.

8. May 1, 2007

cristo

Staff Emeritus
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: $$\frac{d}{dx}e^{Ax}=Ae^{Ax}$$ where A is any constant.

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