Need help with Logistical Derivative Problem

[goatrance]

Homework Statement

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

logistical regression values from ti 83

a=5.71
b=.054
c=5.978

Homework Equations

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

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!

 

[cristo]

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 don't 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!

 

[goatrance]

Thanks a lot cristo. Basicaly I need help factoring this

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

so I can find the second derivative

 

[Integral]

Please do not double (or triple) post. Read Cristo's post. You have not got the 1st derivative correct so the expression you are trying to factor is meaningless.

 

[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.

 

[cristo]

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.

 

[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.

 

[cristo]

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.

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.

but pluging in my equations i get 1.84e^-.054x/(1+ae^-.054x)^2

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