Derivative of P(t) for Equilibrium Point Problem

[pyrosilver]

Homework Statement

I was sick today, and here is what my friend told me. I don't quite understand the question, but apparently we talked about equilibrium points, and for homework, we have to take the derivative of P(t)= 1/((ab^t)+(1/c)), put it in terms of P, and get P'(t)=(P-6)(-k)P.

Homework Equations

The Attempt at a Solution

I'm utterly confused on how to do this.

 

[Liquidxlax]

what are you differentiating with respect to?

 

[Denken]

derivative of P(t)= 1/((ab^t)+(1/c))

well it is in respect to t so if you can find the derivative of b^t than it is -(a(derivative of b^t))/((ab^t)+(1/c))^2 ... assuming that a, b and c are all constants

 

[HallsofIvy]

So
$$P(t)= ((ab)^t+ (1/c))^{-1}$$

Can you differentiate that with respect to t?

It will help to notice that since
$$P(t)= ((ab)^t+ (1/c))^{-1}$$
$1/P= (ab)^t+ 1/c$ so that $(ab)^t= 1/P- 1/c$

 

[Denken]

almost ... it is a(b)^t otherwise agree with that.

 

[pyrosilver]

No, it is (ab)^t, I forgot to specify the a and b going together.

Thanks HallsofIvy! Very helpful :)