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Math Help - Graph tangent equation

  1. #1
    Junior Member Revy's Avatar
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    Exclamation Graph tangent equation

    I have graph x^4+y^4+2=4xy^3 and I need to find tanget equation at (1;1) (on a graph) point.

    I can't seem to get what to do with xs and ys can't convey neither of those

    Do I need to do partial fluxion for x and then for y? (well... these ones: Fx'=... and Fy'=... )

    If yes, what should I do next with those 2 functions?
    Last edited by Revy; March 1st 2010 at 10:58 PM.
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  2. #2
    Member Haven's Avatar
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    For implicit differentiation or Partial Fluxion as you call it. You apply the following:

    1) Take the derivative with respect to x on both sides
    \frac{d}{dx}(x^4+y^4+2) = \frac{d}{dx}(4xy^3)

    2)Any where you see a term with no y's in it, differentiate like normal
    4x^3 + \frac{d}{dx}y^4 = \frac{d}{dx}(4xy^3)

    3) Now in the terms with y's in them, we differentiate with respect to y but keep the term \frac{dy}{dx}. This is essentially the chain rule.
    4x^3 + \frac{dy}{dx}4y^3 = \frac{d}{dx}(4xy^3)

    4) Now in this case we'll have to apply the product rule to the right hand side of the equation
    4x^3 + \frac{dy}{dx}4y^3 = 4y^3 + \frac{dy}{dx}12xy^2

    Now we solve for \frac{dy}{dx}

    \frac{dy}{dx}(4y^3 - 12xy^2) = 4y^3 - 4x^3
    \frac{dy}{dx} = \frac{4y^3 - 4x^3}{4y^3 - 12xy^2}

    Now we substitute in the point (1,1) and we get:
    \frac{dy}{dx} = \frac{4-4}{4-12} = 0
    So the slope of the tangent line at that point is 0.
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  3. #3
    Junior Member Revy's Avatar
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    Tanget equation is just y=1 since it's horizontal for x axis?
    Last edited by Revy; March 2nd 2010 at 12:44 AM.
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