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Math Help - [SOLVED] Implicit Differentiation

  1. #1
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    [SOLVED] Implicit Differentiation

    My teacher didn't really teach this section out of our book that well in class.
    ln(xy) = x + y
    Especially problems with ln(s) and log(s) I tried the problem out and got
    {1+y/-2(xy)^-3}={dy/dx}
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  2. #2
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    \ln(xy)=\ln x+\ln y, now \frac1x+\frac1y\cdot y'=1+y', and now just isolate y'.
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  3. #3
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    Quote Originally Posted by radioheadfan View Post
    My teacher didn't really teach this section out of our book that well in class.
    ln(xy) = x + y
    Especially problems with ln(s) and log(s) I tried the problem out and got
    {1+y/-2(xy)^-3}={dy/dx}
    \ln(xy) = x+y

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

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

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

    y'\left(\frac{1}{y} - 1\right) = 1 - \frac{1}{x}

    y' = \frac{1 - \frac{1}{x}}{\frac{1}{y} - 1}
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  4. #4
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    Quote Originally Posted by radioheadfan View Post
    My teacher didn't really teach this section out of our book that well in class.
    ln(xy) = x + y
    Especially problems with ln(s) and log(s) I tried the problem out and got
    {1+y/-2(xy)^-3}={dy/dx}
    This becomes a lot easier if you first use the fact that \ln(xy) = \ln x+\ln y. Then when you differentiate \ln x+\ln y = x + y implicitly, you get \frac1x + \frac1y\frac{dy}{dx} = 1 + \frac{dy}{dx}. Finally, rearrange that to get an expression for \frac{dy}{dx}.

    Edit. Just beaten to it by skeeter.
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  5. #5
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    You guys are great, I see exactly how this makes sense! Thanks!
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