How to find the first and second derivatives of this logistic function f(t)= L/ (1+Ae^(-kt)).

Question

The question says to assume that L, A, and k are positive real numbers, and that A≥1.

The function is f(t)= L/ (1+Ae^(-kt)), but what is really confusing me is having constants instead of numbers. I don't understand how to take these derivatives ( I know it would use the same rules, I just don't know hw to apply them.)

Answer 1

Let f(t) = L(1+Ae^-kt)^-1

f'(t) = -L(1+Ae^-kt)^-2(Ae^-kt)(-k).  You can combine the -L and -k factors as one constant coefficient kL.

 

 

Note that I used a negative exponent and differentiated by beginning the treatment using the power rule.

Then the chain rule required differentiation of the base which was in parentheses; that is expressed using the fact that the derivative of an exponential is the exponential, and then the chain rule had to be applied again because the exponent of e was not just the variable t, so its derivative gave us the last factor.  Numbers L, A and k are constant factors at various points and those factors apply to the derivatives involved, too.

 

To get f"(t) you will have to use the product rule since there are two factors involving the variable t...and constants will carry through, too.