Find the derivative of the function V(P)= k/P

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SUMMARY

The derivative of the function V(P) = k/P is calculated as V'(P) = -k/P². In the discussion, the user initially substituted k incorrectly as 1 instead of the correct value of k = 8.60, leading to confusion in the derivative calculation. After correcting the substitution, the user arrived at a derivative value of -5.1 when substituting P = 1.30. This highlights the importance of maintaining constant factors in derivative calculations.

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Homework Statement
We are given the function V(P) =k/P. When 𝑉(𝑃) is measured in liters (L) and 𝑃 is measured in atmospheres (atm), the units of 𝑘, a constant based on the identity of the gas, are litres per atmosphere (L/atm). If 𝑘=8.60, what is the instantaneous rate of change of the chamber’s volume with respect to pressure, V'(P) or 𝑑𝑉/𝑑𝑃, when the gas exerts 1.30 atm of pressure on the chamber’s surface? Round your answer to the nearest tenth of a litre per atmosphere.

Find V'(1.30).
Relevant Equations
V(P) =k/P, V'(P)
I tried to find the derivative of the function V(P)= k/P which I found to be:

V'(P) = kP-1 V'(P) = (1)(-1)(P)-1-1 = -1/(P2)

And then I substituted in 1.30 into the derivative to obtain -0.5917 L/atm. And I am kind of confused how to actually find the derivative of this. I thought I was on the right track but I got it wrong, and I don't really know where I would use the k=8.60.
 
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You have lost your constant factor ##k##. For ##f(x)=k\cdot g(x)## we have ##f'(x)=k\cdot g'(x)##. You substituted it by ##k=1##.
 
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fresh_42 said:
You have lost your constant factor ##k##. For ##f(x)=k\cdot g(x)## we have ##f'(x)=k\cdot g'(x)##. You substituted it by ##k=1##.
Okay, thank you. I got -5.1.
 

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