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Math Help - implicit differentiation of log of polynomial

  1. #1
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    Thumbs down implicit differentiation of log of polynomial

    hello.
    find derivative of:
    y = ln(5x + y)
    Using method described at:
    Solutions to Implicit Differentiation Problems
    I get:
    y' = D{ln(5x + y)}
    y' = 1/(5x + y) D{(5x + y)}
    y' = 1/(5x + y) (10x + 2y y')
    y' = 10x/(5x + y) + 2y y'/(5x + y)
    y' - 2y y'/(5x + y) = 10x/(5x + y)
    y'(5x + y)/(5x + y) - 2y y'/(5x + y) = 10x/(5x + y)
    (y'(5x + y) - 2y y')/(5x + y) = 10x/(5x + y)
    y'(5x + y - 2y)/(5x + y) = 10x/(5x + y)
    y' = 10x/(5x + y) (5x + y)/(5x + y - 2y)
    y' = 10x/(5x + y - 2y)

    I like to use online tools to check my work. There are several that purport to solve this type of equation. They all seem to agree that the answer is:
    y' = 10x/(5x + y)
    Of course, they don't show steps, so I can't tell if I'm just wrong or if the online tools are wrong (or if I'm using them wrong).

    Where's the error?
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  2. #2
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    Oct 2009
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    Quote Originally Posted by Jim Pfoss View Post
    hello.
    find derivative of:
    y = ln(5x + y)
    Using method described at:
    Solutions to Implicit Differentiation Problems
    I get:
    y' = D{ln(5x + y)}
    y' = 1/(5x + y) D{(5x + y)}
    y' = 1/(5x + y) (10x + 2y y')
    y' = 10x/(5x + y) + 2y y'/(5x + y)
    y' - 2y y'/(5x + y) = 10x/(5x + y)
    y'(5x + y)/(5x + y) - 2y y'/(5x + y) = 10x/(5x + y)
    (y'(5x + y) - 2y y')/(5x + y) = 10x/(5x + y)
    y'(5x + y - 2y)/(5x + y) = 10x/(5x + y)
    y' = 10x/(5x + y) (5x + y)/(5x + y - 2y)
    y' = 10x/(5x + y - 2y)

    I like to use online tools to check my work. There are several that purport to solve this type of equation. They all seem to agree that the answer is:
    y' = 10x/(5x + y)
    Of course, they don't show steps, so I can't tell if I'm just wrong or if the online tools are wrong (or if I'm using them wrong).

    Where's the error?

    Those online tools suck big time: you are right...and next time trust more your brains and education than lame online "mathematical" tools.

    Tonio
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  3. #3
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    Joined
    Nov 2009
    Posts
    11

    thanks

    Thanks Tonio,

    I may be crazy, but at least I'm not stupid!
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