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Math Help - dy dx of the curve

  1. #1
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    dy dx of the curve

    find dydx of

    x^2 + 3xy + 2y^2 = 5

    My working

    2y^2 = 5 - x^2 - 3xy

    y^2 = 1/2 . (5 - x^2 - 3xy)

    y = root 1/2 . (5 - x^2 - 3xy)

    y = 1/2 . (5 - x^2 - 3xy)^1/2

    y = (2.5 - 1/2x^2 - 1.5xy)^1/2

    y1 = 1/2 (2.5 - 1/2x^2 - 1.5xy)^-1/2. 3

    Advice please if i am close
    Cheers
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  2. #2
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    Quote Originally Posted by Joel View Post
    find dydx of

    x^2 + 3xy + 2y^2 = 5

    My working

    2y^2 = 5 - x^2 - 3xy

    y^2 = 1/2 . (5 - x^2 - 3xy)

    y = root 1/2 . (5 - x^2 - 3xy)

    y = 1/2 . (5 - x^2 - 3xy)^1/2

    y = (2.5 - 1/2x^2 - 1.5xy)^1/2

    y1 = 1/2 (2.5 - 1/2x^2 - 1.5xy)^-1/2. 3

    Advice please if i am close
    Cheers
    first of all, it's dy/dx , not dydx.

    second, you're not even close in your "method".

    research finding implicit derivatives in your text or on the web.
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  3. #3
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    Just to clarify a bit on what skeeter is saying:

    To use the more "traditional" method of solving for a derivative, you would need to solve for y explicitly. What that means is that you can get the equation into the form y=\cdots where there are no y terms on the right hand side at all.

    Although your work to solve for y seemed like it was on the right track, the problem is you are stuck with a y term in the right hand side.

    The correct way to handle this problem is to use implicit differentiation. This method is used without first solving for y.

    Just to give you a small hint -- You do not need to simply or modify the equation at all before taking the derivative! Have you learned implicit differentiation techniques?
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  4. #4
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    Quote Originally Posted by Joel View Post
    find dydx of

    x^2 + 3xy + 2y^2 = 5

    My working

    2y^2 = 5 - x^2 - 3xy

    y^2 = 1/2 . (5 - x^2 - 3xy)

    y = root 1/2 . (5 - x^2 - 3xy)

    y = 1/2 . (5 - x^2 - 3xy)^1/2

    y = (2.5 - 1/2x^2 - 1.5xy)^1/2

    y1 = 1/2 (2.5 - 1/2x^2 - 1.5xy)^-1/2. 3

    Advice please if i am close
    Cheers
    When you have f(y), it is the same as evaluating f(x) except that there is a \frac{dy}{dx} attached (multiplied).

    So deriving y^2 = x gives

    2y\frac{dy}{dx} = 1

    \frac{dy}{dx} = \frac{1}{2y}

    This is simply a special case of the chain rule: notice that when evaluating x's this rule is hidden by the fact \frac{dx}{dx} = 1.

    Product rules with f(y)g(x) work as you would expect. Can you use this technique to answer your question?
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