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

May 2010
4
0
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!
 
Last edited by a moderator:
Jan 2010
354
173
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:

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

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

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