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

[TsAmE]

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

 

[rock.freak667]

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.

 

[TsAmE]

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 :(

 

[System]

[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 ..