Differential change of variables

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Homework Help Overview

The discussion revolves around a differential equation involving partial derivatives, specifically the expression ∂P/∂r = (1/r)∂ψ/∂r - ψ/r^2. The original poster seeks clarification on the origin of the term -ψ/r^2 and its application to higher order differentials.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are examining the relationship between the variables P, ψ, and r, and discussing the differentiation process involved in deriving the expression. There are questions about the specific rules or principles that lead to the inclusion of the term -ψ/r^2.

Discussion Status

Some participants are providing insights into the differentiation process, while others are seeking further clarification on the relationships between the variables involved. The discussion is ongoing, with multiple interpretations being explored.

Contextual Notes

There is a noted lack of clarity regarding the definitions and relationships of the variables P and ψ, which may impact the understanding of the problem. The original poster has acknowledged a potential misunderstanding in the notation used.

sjw11
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I am working on an exam question, I have the solution but i can't figure out a step in between,

I need to show that ∂P/∂r = (1/r)∂ψ/∂r - ψ/r^2

I am wondering where the - ψ/r^2 comes from. I.e. which rule is this, and sencondly how does this apply to higher order differentials.

Thanks in advance for all your help

Stuart
 
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sjw11 said:
I am working on an exam question, I have the solution but i can't figure out a step in between,

I need to show that ∂P/∂r = (1/r)∂ψ/∂r - ψ/r^2

I am wondering where the - ψ/r^2 comes from. I.e. which rule is this, and sencondly how does this apply to higher order differentials.

Thanks in advance for all your help

Stuart
Hello Stuart. Welcome to PF !

Without knowing how P, ψ, and r are related, it will be hard to answer your question.

Is that P supposed to be ρ ?
 
Sammy,

Apologies I forgot to put that in

ψ=rP

p=P not ρ

cheers
 
Then P = [itex]\frac{ψ}{r}[/itex]

[tex]\frac{∂P}{∂r} = \frac{∂}{∂r}(ψ\cdot\frac{1}{r}) = \frac{1}{r}\cdot\frac{∂ψ}{∂r} + ψ\cdot\frac{∂}{∂r}(\frac{1}{r})[/tex]

...
 

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