# D/dx( ln( x^2 + y^2 ) )

1. Apr 19, 2010

### TsAmE

1. The problem statement, all variables and given/known data

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

2. Relevant equations

d / dx ( lnx ) = 1 / x

3. The attempt at a solution

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

I couldnt find a way to isolate y' on its own

2. Apr 19, 2010

### rock.freak667

Cross multiply by x2+y2

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

Should be easier to simplify now.

3. Apr 20, 2010

### 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 :(

4. Apr 20, 2010

### System

$$y' (x^2+y^2) = 2x+2yy'$$

$$y'(x^2+y^2)-2yy'=2x$$

take y' a common factor and complete ..