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    Can any1 solve this question 4 me cuz i have a C4 paper after like 16 Hrs.
    x^3-2xy-4x+y^3-51=0
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    (Original post by Knight91)
    Can any1 solve this question 4 me cuz i have a C4 paper after like 16 Hrs.
    x^3-2xy-4x+y^3-51=0
    I don't think this is the whole question. You have two unknowns so you need two equations.
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    sry abt dat u have to find dy/dx at point (4,3)
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    (Original post by Knight91)
    sry abt dat u have to find dy/dx at point (4,3)
    Implicit differentiate it, then plug the values in
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    yeh i do know da method but i'm not getting da same answer as the mark scheme!!!!so can any1 do it step by step cuz i'm not getting it from da mark scheme!!
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    the simple rule for implicit diff, is diff x's normally, and when you encounter a y, diff that normally but stick dy/dx on to it. eg y^2 = 2y(dy/dx)
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    (Original post by Knight91)
    yeh i do know da method but i'm not getting da same answer as the mark scheme!!!!so can any1 do it step by step cuz i'm not getting it from da mark scheme!!
    Show me what you have done so far, so i can see where you went wrong.
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    x^3-2xy-4x+y^3-51=0

    3x^2 - 2[xdy/dx + y] + 3y^2 = 0

    can you get it from here?
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    yeah i actually got till dy/dx=(4-2y-3x^2)/(-2x+3y^2) I go wrong after putting in the values.....
    i'm getting -50/19 whereas the answer is -38/19
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    If there are two unknowns AND they are multiplied together, I'm not sure that I can do it or even suggest what to do. You must be missing something?
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    (Original post by play_fetch)
    You must be missing something?
    :nah:

    x^3 - 2xy - 4x + y^3 - 51 = 0

    3x^2 - 2x\frac {dy}{dx} - 2y - 4 + 3y^2\frac {dy}{dx} = 0

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

    \frac {dy}{dx} = \frac {4 +2y - 3x^2}{3y^2 - 2x}
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    u r absolutely rite till here now put (4,3) in this equation to find the exact value of dy/dx....
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    hey thanx man i actually got my mistake insted of 4+2Y I had written 4-2Y now i'm getting the same answer as the markscheme...thanx again helped alot...
 
 
 
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