Taking Differentials to Find Partial Derivative

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Homework Help Overview

The discussion revolves around finding the partial derivative of pressure \( p \) with respect to volume \( V \) from the equation \((p + \frac{a}{{V^2 }})(V - b) = C\) using differentials. Participants are exploring the constraints of the problem, particularly the requirement to use differentials rather than rearranging the equation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply differentials to the equation but expresses uncertainty about handling the brackets without expanding them. Some participants suggest using the product rule for differentials, while others question how to apply it effectively without leading to complications.

Discussion Status

Participants are actively discussing the application of the product rule and its implications. There is a recognition of the challenge posed by the requirement to not rearrange the equation, and some guidance has been offered regarding the use of the product rule. However, there is no explicit consensus on the best approach to take.

Contextual Notes

There is a specific constraint mentioned regarding the inability to rearrange the equation, which is causing confusion among participants. The original poster indicates that they must adhere to this constraint while attempting to find the partial derivative.

jamjar
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Hi,
I'm trying to take differentials of the following equation

[tex](p + \frac{a}{{V^2 }})(V - b) = C[/tex]

in order to find the partial derivative [tex]\frac{{\partial p}}{{\partial V}}[/tex]

I know there's an easier way to do it but I have to take differentials.
I'm just not sure how to deal with the brackets without multiplying out (I can't rearrange the equation).

Any hints welcome.
 
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Use the product rule!
 
How do I do that with differentials?
I end up with crazy results.
 
You "can't" rearrange the equation, as in the problem won't let you? The product rule for differentials is [tex]d(fg) = g df + f dg[/tex].
 
If I use that product rule I end up with a free floating [tex]p[/tex] in my equation where I know that [tex]\frac{{\partial p}}{{\partial V}}[/tex] doesn't have a term in [tex]p[/tex]
 
Yes, you then have to solve for p in terms of V using the original equation. This I why I don't undestand why you can't just solve for p in terms of V from the start since you have do it eventually anyway.
 

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