How can y' be isolated in y'(x^2 + y^2) - 2yy' = 2x?

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Homework Help Overview

The problem involves isolating the derivative y' in the context of the equation y' (x^2 + y^2) - 2yy' = 2x, where y is defined as y = ln(x^2 + y^2). The subject area pertains to implicit differentiation and algebraic manipulation of derivatives.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss attempts to isolate y' from the equation, with some expressing difficulty in achieving this goal. There are references to cross-multiplying and simplifying expressions involving y' and other terms.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and reasoning. Some have suggested methods for simplification, but there is no clear consensus on how to isolate y' effectively. The conversation reflects a collaborative exploration of the problem.

Contextual Notes

Participants are working under the constraints of implicit differentiation and the specific functional form of y. There is an acknowledgment of the complexity involved in isolating y' due to its presence in multiple terms.

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Homework Statement



Find y' if y = ln( x^2 + y^2 )

Homework Equations



d / dx ( lnx ) = 1 / x

The Attempt at a Solution



y' = ( 1 / ( x^2 + y^2 ) ) * 2x + 2y * y'
( y' / 2x + 2y * y' ) = ( x^2 + y^2 )^-1

I couldn't find a way to isolate y' on its own
 
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TsAmE said:

The Attempt at a Solution



y' = ( 1 / ( x^2 + y^2 ) ) * 2x + 2y * y'
( y' / 2x + 2y * y' ) = ( x^2 + y^2 )^-1

Cross multiply by x2+y2


y'(x2+y2)=2x+2y*y'

Should be easier to simplify now.
 
When I cross multiply y' / (2x + 2y * y') = (x^2 + y^2)^-1 I get:

y' = (x^2 + y^2)^-1(2x + 2y * y') but the y' s arent isolated :(
 
[tex]y' (x^2+y^2) = 2x+2yy'[/tex]

[tex]y'(x^2+y^2)-2yy'=2x[/tex]

take y' a common factor and complete ..
 

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