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Thread: Implicit Differentiation

  1. #1
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    Implicit Differentiation

    Find dy/dx

    $\displaystyle y^2=\frac{x-1}{x=1}$

    I got $\displaystyle \frac{4y}{x^2+x+1}$

    The answer is $\displaystyle \frac{1}{y(x+1)^2}$

    I believe I messed up with the fraction over a number, as in (what is the LaTex for this? It's not in the tutorial) 2/x^2+x+1/2y. I don't know how to deal with this. For some reason, perhaps a habit, I want to bring 2y to the top and multiply it by 2. After seeing the answer, I think it's the top times $\displaystyle \frac{1}{2y}$.

    What is it? and if I'm right about the mistake? How come you do that?
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  2. #2
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    I just understood your question. Yes when you divide the 2y it goes on the bottom not the top. Think back to dividing simple fractions. A fraction divided by a fraction you take the reciprocal of the second and multiply. ie

    $\displaystyle \frac{\frac{1}{2}}{3}=(\frac{1}{2})(\frac{1}{3})$

    I assume the denominator is x+1

    Using the quotient rule on the right side

    $\displaystyle 2y\frac{dy}{dx}=\frac{(x+1)(1)-(x-1)(1)}{(x+1)^2}$

    Simplify numerator: $\displaystyle 2y\frac{dy}{dx}=\frac{2}{(x+1)^2}$
    Solve for dy/dx:$\displaystyle \frac{dy}{dx}=\frac{2}{2y(x+1)^2}$
    Simplify: $\displaystyle \frac{dy}{dx}=\frac{1}{y(x+1)^2}$
    Last edited by jabroni1212; Nov 29th 2007 at 04:57 PM.
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  3. #3
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    Quote Originally Posted by Truthbetold View Post
    Find dy/dx

    $\displaystyle y^2=\frac{x-1}{x+1}$
    $\displaystyle \frac{{x - 1}}
    {{x + 1}} = \frac{{x + 1 - 2}}
    {{x + 1}} = 1 - \frac{2}
    {{x + 1}}.$

    It takes less time to differentiate. (You don't even need product rule or quotient rule.)
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