# Thread: Need help finding the derivative of y = ln(x^2 + y^2)

1. ## Need help finding the derivative of y = ln(x^2 + y^2)

I know I need to use implicit differentiation.... but I can't seem to solve for y' when y = ln(x^2 + y^2) Please help me, I'd really appreciate it!

2. Originally Posted by ktehspynx
I know I need to use implicit differentiation.... but I can't seem to solve for y' Please help me, I'd really appreciate it!
One method you can use is to first get rid of the logarithm:

$\displaystyle y=\ln(x^2+y^2) \implies e^y = x^2+y^2$

This is somewhat easier to differentiate, but the original way works too:

$\displaystyle y = \ln(x^2 + y^2)$

$\displaystyle \implies y' = \frac{\frac{d}{dx}[x^2+y^2]}{x^2+y^2} = \frac{ 2x + 2y y' }{x^2+y^2}$

To solve for $\displaystyle y'$ we need to multiply the denominator over first:

$\displaystyle \implies (x^2+y^2)y' = 2x + 2y y'$

$\displaystyle \implies (x^2+y^2-2y)y' = 2x$

$\displaystyle \implies y' = \frac{2x}{x^2+y^2-2y}$

3. Thanks so much, this helped a lot.