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

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In summary, the conversation discussed finding the derivative of y = ln(x^2 + y^2) using the equation d/dx (lnx) = 1/x. The attempt at a solution involved using the chain rule to simplify the equation, but the y' could not be isolated. A suggestion was made to cross multiply to isolate y', resulting in the equation y'(x^2 + y^2) - 2yy' = 2x. From there, the conversation ended with the suggestion of taking y' as a common factor and completing the equation to find the final derivative.
  • #1
TsAmE
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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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  • #2
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.
 
  • #3
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
[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 ..
 

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

What is the derivative of ln( x^2 + y^2 ) with respect to x?

The derivative of ln( x^2 + y^2 ) with respect to x is 2x/(x^2 + y^2).

Is the derivative of ln( x^2 + y^2 ) with respect to y equal to zero?

Yes, the derivative of ln( x^2 + y^2 ) with respect to y is zero as y is not present in the expression.

Can the chain rule be used to find the derivative of ln( x^2 + y^2 )?

Yes, the chain rule can be used to find the derivative of ln( x^2 + y^2 ).

Is the derivative of ln( x^2 + y^2 ) continuous?

Yes, the derivative of ln( x^2 + y^2 ) is continuous as ln( x^2 + y^2 ) is a differentiable function.

What is the graph of the derivative of ln( x^2 + y^2 )?

The graph of the derivative of ln( x^2 + y^2 ) is a curve that approaches zero at x = 0 and y = 0 and becomes steeper as x and y increase.

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