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Derivative Problem

  1. Jun 3, 2004 #1
    [tex]e^{xy} = ln (x+y)[/tex]

    I need to find dy/dx...but its difficult to get the answer in the book

    I tried this:

    [tex]ln e^{xy} = ln (ln (x + y))[/tex]

    [tex]xy = ln (ln (x+y))[/tex]

    taking the derivitive in terms of x

    [tex]y + \frac{dy}{dx}x = (\frac {1}{ln (x+y)})(\frac {1}{x+y})(1 + \frac {dy}{dx})[/tex]

    If I were to continue and solve for dy/dx It would not even be close to the books answer. I'm probably doing it wrong anyways...can someone show me the right way?
  2. jcsd
  3. Jun 3, 2004 #2


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    Science Advisor
    Homework Helper

    Try implicitly differentiating right off the bat. I find sometimes with implicit differentiation questions, your answers and the books answers can often seem pretty different (when in fact being the same).

    [tex]e^{xy} = ln(x + y)[/tex]
    [tex]e^{xy}(y + xy') = \frac{1 + y'}{x + y}[/tex]
    [tex]y' = \frac{(x + y)(y)(e^{xy}) - 1}{1 - xe^{xy}}[/tex]

    If you can find some way to further simplify, go ahead.
  4. Jun 3, 2004 #3
    Hey, thanks :)
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