Derivative of P(t) for Equilibrium Point Problem

pyrosilver
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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.
 
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what are you differentiating with respect to?
 
pyrosilver said:
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
 
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
 
almost ... it is a(b)^t otherwise agree with that.
 
No, it is (ab)^t, I forgot to specify the a and b going together.

Thanks HallsofIvy! Very helpful :)
 
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