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

  1. #1
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    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!
    Last edited by mr fantastic; May 27th 2010 at 07:40 PM. Reason: Copied expression from post titled.
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  2. #2
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    Quote Originally Posted by ktehspynx View Post
    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:

    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:

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

    \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 y' we need to multiply the denominator over first:

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

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

    \implies y' = \frac{2x}{x^2+y^2-2y}
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  3. #3
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    Thanks so much, this helped a lot.
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