Derivative of a logistic function

I'm having a little trouble figuring out how to calculate the derivative of a logistic function.

$\displaystyle {\left(\frac{3000}{1 + 9e^{-0.4055t}}\right)} $

I'm starting to feel like this 'introduction to calculus' is a little bit above me, but oh well.

I originally just used the quotient rule, $\displaystyle {\left(\frac{g(t)f '(t) - f(t)g '(t)}{(g(t))^2}\right)}$

with $\displaystyle f(t)=3000... f '(t)= 0.... g(t)= 1 + 9e^{-0.4055t}.... g '(t)= 3.6495e^{-0.4055t}$

Using those figures there, I ended up with

$\displaystyle {\left(\frac{0 - 29,192}{7.71}\right)}$ when solving for t=4.

Where am I going wrong?

Re: Derivative of a logistic function

Quote:

Originally Posted by

**astuart** I'm having a little trouble figuring out how to calculate the derivative of a logistic function.

$\displaystyle {\left(\frac{3000}{1 + 9e^{-0.4055t}}\right)} $

I'm starting to feel like this 'introduction to calculus' is a little bit above me, but oh well.

I originally just used the quotient rule, $\displaystyle {\left(\frac{g(t)f '(t) - f(t)g '(t)}{(g(t))^2}\right)}$

with $\displaystyle f(t)=3000... f '(t)= 0.... g(t)= 1 + 9e^{-0.4055t}.... g '(t)= 3.6495e^{-0.4055t}$

Using those figures there, I ended up with

$\displaystyle {\left(\frac{0 - 29,192}{7.71}\right)}$ when solving for t=4.

Where am I going wrong?

1. g'(t) must be negative.

2. $\displaystyle (0-3000 \cdot (-3.6495 \cdot e^{-0.4055 t})) \approx 2162.365$

So check your calculations of the numerator.

Re: Derivative of a logistic function

Quote:

Originally Posted by

**earboth** 1. g'(t) must be negative.

2. $\displaystyle (0-3000 \cdot (-3.6495 \cdot e^{-0.4055 t})) \approx 2162.365$

So check your calculations of the numerator.

Damn, another simple mistake. Math would have to be the most frustrating yet satisfying subjects. A little error can throw you out so much and unless it's picked up, it doesn't matter how the problem is set up, or what formulas you use, you won't get the correct answer..

$\displaystyle d/dt = {\left(\frac{3000}{1 + 9e^{-0.4055t}}\right)}

= {\left(\frac{(0*3000) - (-3.6495e^{-0.4055t})}{(1 + 9e^{-0.4055t})^2}\right)}$

Plugging in t=4 gets me 280, which is what i was after.

Thank you!!!