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Math Help - can someone explain this

  1. #1
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    can someone explain this

    i have a problem that says find dy/dx: ln(xy)=x+y. am i trying to find the derv of both sides? dont they equal the same thing?[IMG]file:///C:/Users/Johnny/AppData/Local/Temp/moz-screenshot-1.png[/IMG][IMG]file:///C:/Users/Johnny/AppData/Local/Temp/moz-screenshot.png[/IMG]
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  2. #2
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    Quote Originally Posted by johnnyc418 View Post
    i have a problem that says find dy/dx: ln(xy)=x+y. am i trying to find the derv of both sides? dont they equal the same thing?[IMG]file:///C:/Users/Johnny/AppData/Local/Temp/moz-screenshot-1.png[/IMG][IMG]file:///C:/Users/Johnny/AppData/Local/Temp/moz-screenshot.png[/IMG]
    note that \ln(xy) = \ln(x) + \ln(y)

    \frac{d}{dx} \left[\ln(x) + \ln(y) = x+y\right]

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

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

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

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

    \frac{y - xy}{xy - x} = y'
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  3. #3
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    Quote Originally Posted by johnnyc418 View Post
    i have a problem that says find dy/dx: ln(xy)=x+y. am i trying to find the derv of both sides? dont they equal the same thing?

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

    So,

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

    Let's differentiate both sides implicitly,

    \frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1+\frac{dy}{d  x}

    And now move stuff around and solve for \frac{dy}{dx}
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