Need help with simple derivative

[gilroy222]

Hello. I am new to the forums. I figured this would be a great place to get help as I generally struggle with any sort of algebra (which calculus is based upon).

I am trying to do the derivative of 500/(1+e^-t)

The best I can do is 500(1+e^-t)^-1=500(-1)(1+e^-t)^-2*e^-t but then I get stuck. I am not sure exactly how to keep going.
If someone could explain how you determine what steps to take next I would be truely greatful!

I am doing a self-study calculus course in prep for university and unfortuneatly they do not have a tutoring or a help service in which you can ask a teacher.

Thanks

 

[Deleted member 4993]

Hello. I am new to the forums. I figured this would be a great place to get help as I generally struggle with any sort of algebra (which calculus is based upon).

I am trying to do the derivative of 500/(1+e^-t)

The best I can do is 500(1+e^-t)^-1=500(-1)(1+e^-t)^-2*e^-t but then I get stuck. I am not sure exactly how to keep going.
If someone could explain how you determine what steps to take next I would be truely greatful!

I am doing a self-study calculus course in prep for university and unfortuneatly they do not have a tutoring or a help service in which you can ask a teacher.

Thanks

Do you know the quotient rule of derivative?

You need to use that.

If you are taking calculus for the first time - it is NOT a very good idea to take it on-line.

 

[gilroy222]

Do you know the quotient rule of derivative?

You need to use that.

If you are taking calculus for the first time - it is NOT a very good idea to take it on-line.

I forgot about the quotient rule! Thanks a lot!

I agree, I really did not want to take it online because I learn much better in a classroom. However, because of my co-op I was unable to do it through my local Adult Learning Centre in time before I begin University.

 

[gilroy222]

Now I am not sure what exactly I am doing wrong here, but using the quotient rule I get to

(1+e^-t)-500(e^-t) / (1+e^-t)^2

I did some googling and it seems the answer is supposed to be

500(e^-t)/(1+e^-t)^2

So it seems like I am really close, just I am missing something small and cannot quite put my finger on it.

I should add that it is a population problem in which t = time in years and I will have to plug in 3 years.

Thanks

 

[Deleted member 4993]

derivative of 500/(1+e^-t)

Thanks

This problem can be solved in several "equivalent" ways to get the same answer

d/dt(1+ e^-t) = (-1) e^-t= - e^-t

d/dt [500/(1+e^-t)] = 500[-1/(1+e^-t)²] * [- e^-t] = [500*e^-t/(1+e^-t)].....by chain rule

d/dt [500/(1+e^-t)] = [0*(1+e^-t) - 500*(- e^-t)]/(1+e^-t)² = [500*e^-t/(1+e^-t)].....by quotient rule rule

 

[gilroy222]

d/dt(1- e^-t) = - (-1) e^-t= e^-t

d/dt [500/(1-e^-t)] = 500[e^-t/(1-e^-t)²]

I am sorry, I am not quite understanding exactly why you have (1-e...)

Forgive me for asking, but I noticed your last name....are you the gentleman who runs Khan Academy??

Thanks

 

[gilroy222]

I believe I have found the answer. http://www.freemathhelp.com/forum/t...ves-involving-e-deriv-of-500-divided-by-1-e-t

Found that old thread and attempted to try the chain rule again. I achieved the same answer.

Just thought I would put that there to help anyone reading this in the future. I am curious as to why in this case we must use chain rule, when normally, as Khan pointed out, it should be the quotient rule.

 

[Ishuda]

...
The best I can do is the derivative of 500(1+e^-t)^-1=500(-1)(1+e^-t)^-2*e^-t but then I get stuck. I am not sure exactly how to keep going....

See correction in red. Almost. First of all what is the derivative of 1 + e^-t. It's not quite e^-t. But other than that, you're done:
\(\displaystyle \frac{d(500 (1 + e^{-t})^{-1})}{dt} = \frac{-500 (1 + e^{-t})^{-2}\frac{d(1 + e^{-t})}{dt}}{dt}\)