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Math Help - More implicit differentiation :(

  1. #1
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    More implicit differentiation :(

    ln(xy) = 2 - x - y

    The answer to this question in my book is dy/dx = [(-x - 1)/x][y/(1 + y)]

    Could someone show me how to approach this question? I know I have to differentiate the ln(xy) term and apply the product rule to the (xy) inside, but I don't know how to do that exactly.

    Thank!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by samstark View Post
    ln(xy) = 2 - x - y

    The answer to this question in my book is dy/dx = [(-x - 1)/x][y/(1 + y)]

    Could someone show me how to approach this question? I know I have to differentiate the ln(xy) term and apply the product rule to the (xy) inside, but I don't know how to do that exactly.

    Thank!
    Remember that the chain rule is also involved here. So you need to consider:
    \frac{d}{dx} f(g(x)) = \frac{df}{dg} \cdot \frac{dg}{dx}

    In this case f(u) = ln(u) and u = xy.
    \frac{d}{du} ln(u)  = \frac{1}{u}

    d(xy) = y + x \frac{dy}{dx}

    See where this gets you.

    -Dan
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  3. #3
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    e^(i*pi)'s Avatar
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    You could also use the log laws on the left hand side

    \ln(ab) = \ln(a) + \ln(b)
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