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Math Help - implicit differentiation

  1. #1
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    implicit differentiation

    Can anyone show me how to find \frac{dy}{dx} for (x^2+y^2-1)(y-x^2-1) = 0 at the points (0,1) and (0,0).

    I know that I need to use the process of implicit differentiation but I don't know exactly how to apply it here. Do I need to expand the brackets first and then differentiate everything?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by demode View Post
    Can anyone show me how to find \frac{dy}{dx} for (x^2+y^2-1)(y-x^2-1) = 0 at the points (0,1) and (0,0).

    I know that I need to use the process of implicit differentiation but I don't know exactly how to apply it here. Do I need to expand the brackets first and then differentiate everything?
    you don't have to expand the brackets. just use the product rule. remember, when differentiating a y-term, you need to attach dy/dx to it.
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  3. #3
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    JHevon brings up a good point, but if you DO expand the brackets, several terms cancel, thus making it easier to calculate the derivative.

    (x^2 + y^2 - 1)(y - x^2 - 1) = 0

    x^2(y - x^2- 1) + y^2(y - x^2 - 1) - 1(y - x^2 - 1) = 0

    x^2y - x^4 - x^2 + y^3 - x^2y - y^2 - y + x^2 + 1 = 0

    -x^4 + y^3 - y^2 - y + 1 = 0

    y^3 - y^2 - y + 1 = x^4

    \frac{d}{dx}(y^3 - y^2 - y + 1) = \frac{d}{dx}(x^4)

    \frac{d}{dy}(y^3 - y^2 - y + 1)\frac{dy}{dx} = 4x^3

    (3y^2 - 2y - 1)\frac{dy}{dx} = 4x^3

    (3y + 1)(y - 1)\frac{dy}{dx} = 4x^3

    \frac{dy}{dx} = \frac{4x^3}{(3y + 1)(y - 1)}.
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