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Math Help - implicit differentiation

  1. #1
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    implicit differentiation

    I don't really understand what to do for implicit differentiation. For example, how do I go about doing this:

    Find dy/dx by implicit differentiation. 4x2 + 4xy - y2 = 1
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by kwivo View Post
    I don't really understand what to do for implicit differentiation. For example, how do I go about doing this:

    Find dy/dx by implicit differentiation. 4x2 + 4xy - y2 = 1
    Implicit differentiation is really just the chain rule in disguise.

    If you have a function f(y) and take y = y(x) then you've got f(y(x)). Take the x derivative and you get f'(y(x)) y'(x) by the chain rule. This is exactly the same thing as saying
    \frac{d}{dx}f(y) = f^{\prime}(y) \cdot \frac{dy}{dx}

    So let's take your problem:
    4x^2 + 4xy - y^2 = 1

    Take the derivative:
    4 \cdot 2x + \left ( 4y + 4x \frac{dy}{dx} \right ) - 2y \cdot \frac{dy}{dx} = 0

    Now solve for \frac{dy}{dx}

    -Dan
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kwivo View Post
    I don't really understand what to do for implicit differentiation. For example, how do I go about doing this:

    Find dy/dx by implicit differentiation. 4x2 + 4xy - y2 = 1
    for implicit differentiation, you just find the derivative of each variable, taking account for what you are differentiating and with respect to what.

    if you differentiate an x-term you attach dx/dx to it (since you took the derivative of the variable x with respect to x), we usually don't write it though, because derivative notations can function as fractions and so dx/dx cancels to become 1

    similarly, when differentiating a y-term, we attach dy/dx to it, since we took the derivative of y with respect to x. we often write y' when it is understood what we are differentiating with respect to

    here goes:


    4x^2 + 4xy - y^2 = 1

    differentiating implicitly, we get:

    8x + 4y + 4x~y' - 2y~y' = 0

    now solve for y', which is \frac {dy}{dx}



    EDIT: Beaten by "the man." And he gave a better explanation too!
    Last edited by Jhevon; October 8th 2007 at 02:16 PM.
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  4. #4
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    One more question: I don't understand how you get the dy/dx after the 2y. Where did that come from? Isn't 2y already the derivative of y^2.
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    is up to his old tricks again! Jhevon's Avatar
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    did you read the posts? what did you not get?
    Quote Originally Posted by topsquark View Post
    If you have a function f(y) and take y = y(x) then you've got f(y(x)). Take the x derivative and you get f'(y(x)) y'(x) by the chain rule. This is exactly the same thing as saying
    \frac{d}{dx}f(y) = f^{\prime}(y) \cdot {\color{red}\frac{dy}{dx}}
    Quote Originally Posted by Jhevon View Post
    for implicit differentiation, you just find the derivative of each variable, taking account for what you are differentiating and with respect to what.

    when differentiating a y-term, we attach dy/dx to it, since we took the derivative of y with respect to x.
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  6. #6
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    Sorry, I overlooked that part.
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