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Math Help - Partial differentiation ln(x+y)=x

  1. #1
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    Partial differentiation ln(x+y)=x

    Hello guys,
    wondering if anyone can help me out with this problem.
    find dy/dx:
    ln(x+y)=x

    why is this solution invalid:
    ln(x+y)=x
    ex=x+y
    y=ex-x
    (dy/dx)=ex-1

    Many thanks.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Partial differentiation ln(x+y)=x

    I believe you meant to title your topic "Implicit differentiation..."

    Your result is valid, it is just in a different form than that which is obtained by implicit differentiation.

    Given \ln(x+y)=x then implicit differentiation yields:

    \frac{1+y'}{x+y}=1

    y'=x+y-1

    Now, if we substitute for y, we then find:

    y'=x+e^x-x-1=e^x-1
    Thanks from topsquark
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