Partial Derivatives of a and b

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Homework Help Overview

The discussion revolves around finding the partial derivatives of variables a and b, given the equations m = a + b and n = a² + b². Participants are attempting to clarify their understanding of the relationships between these variables and the process of differentiation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are questioning the correctness of their initial attempts at finding the partial derivatives, particularly (dm/db)a and (db/dm)n. There is a discussion about expressing b in terms of n and m, and how to differentiate accordingly. Some participants express uncertainty about their methods and whether implicit differentiation is appropriate.

Discussion Status

Some guidance has been offered regarding the differentiation process, suggesting that participants should express b as a function of n and m before differentiating. There is an ongoing exploration of different interpretations of the problem, but no explicit consensus has been reached.

Contextual Notes

Participants are navigating the complexities of implicit differentiation and the relationships between the variables, with some expressing uncertainty about the steps taken and the correctness of their approaches.

Liquidxlax
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Homework Statement




m=a+b n=a2+b2

find partials (dm/db)a and (db/dm)n




The Attempt at a Solution




(dm/db)a = 1 is that right?

and

(db/dm)n I'm not sure how to get all the variables into one equation but

a = sqrt(b2-n)

so

m = b - sqrt(b2-n)

can someone help out please?
 
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Liquidxlax said:

Homework Statement




m=a+b n=a2+b2

Wait, so are you saying that m = a + bn AND m = a2 + b2?
 
cepheid said:
Wait, so are you saying that m = a + bn AND m = a2 + b2?

no

m=a+b and n=a^2 +b^2
 
Liquidxlax said:
no

m=a+b and n=a^2 +b^2

Okay.

Liquidxlax said:
(dm/db)a = 1 is that right?

Yeah.


Liquidxlax said:
and

(db/dm)n

This suggests that you want the function b(n,m), so you have to take what you have

Liquidxlax said:
m = b - sqrt(b2-n)

solve it for b, and then differentiate that w.r.t. m, keeping n constant.
 
cepheid said:
Okay.



Yeah.




This suggests that you want the function b(n,m), so you have to take what you have



solve it for b, and then differentiate that w.r.t. m, keeping n constant.


well i came here to late, i did try to do it the way you stated, but I'm not sure if i did it right. I could have done it by implicit differentiation or w.e it is called, which i did do. So we shall see whether i got it or not.
 

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