Question

Derivative of P?? P(t)= A / 1+Be^(-kt). A, B, k are positive constants.

Answer 1

Is this your function?\[P(t)=\frac{A}{1+Be^{-kt}}\]

Answer 2

yesss

Answer 3

first, re-write it as:\[P(t)=A(1+Be^{-kt})^{-1}\]

Answer 4

Now let u=1+Be^(-kt) so du=-Bke^(-kt)dt...ok so far?

Answer 5

okay

Answer 6

Now we have \[P(t)=Au^{-1}\]So taking derivative, using the chain rule, we get:\[dP/dt=(dP/du)(du/dt)\]or,\[dP/dt=-Au^{-2}(-Bke^{-kt})\]Getting rid of our new variable "u" we get:\[dP/dt=-A(1+Be^{-kt})^{-2}(-Bke^{-kt})\]I'll leave it for you to simplify if you wish.

Answer 7

sorry, first line shoud really say P(u)=Au^(-1) as it is a function of u now, not t

Answer 8

gotta run my little ones to daycare...I'll check back with you in a bit if you have questions on this one

Answer 9

okayyy thanks!

Answer 10

okay i got it it makes a lot more sense now.

Answer 11

would you know how to evaluate the limit of P(t) toward infinity and -infinity?

Answer 12

plug it in for t but I dont know what that gets me? 1+Be^-kinfinity?

Answer 13

when it approaches infinity I got A/1 or A which is correct. However I keep getting it wrong when it approaches - infinity