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    Write down grad ar point (1,1) at curve
    y=2/(3-x^2)

    Maybe I am really tired and being stupid but...if i use quotient rule so..

    u=2 => du/dx=???? ...do i just put zero?

    It also says write down so it seems to be implying its v easy...i am being daft and missing something..ahhh help my tired brain!!
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    (Original post by shiny)
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    nope, answer says 1
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    (Original post by BossLady)
    Write down grad ar point (1,1) at curve
    y=2/(3-x^2)

    Maybe I am really tired and being stupid but...if i use quotient rule so..

    u=2 => du/dx=???? ...do i just put zero?

    It also says write down so it seems to be implying its v easy...i am being daft and missing something..ahhh help my tired brain!!
    You could take the constant 2 outside the diiferentiation so you have y = 2(1/3-x²), and differentiate the bracket...
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    (Original post by ZJuwelH)
    You could take the constant 2 outside the diiferentiation so you have y = 2(1/3-x²), and differentiate the bracket...
    lol, ZJuwelH, you're a star, and I have to stop doing maths after 10pm
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    (Original post by BossLady)
    lol, ZJuwelH, you're a star, and I have to stop doing maths after 10pm
    I could do maths all day (please no external force ever hold me to that)
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    (Original post by BossLady)
    nope, answer says 1
    Oops ... missed the square!

    Yes its 1

    dy/dx = d/dx { 2 / (3 - x^2) } = 4x / (3 - x^2)^2

    At x = 1; dy/dx = 1!
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    (Original post by BossLady)
    Write down grad ar point (1,1) at curve
    y=2/(3-x^2)

    Maybe I am really tired and being stupid but...if i use quotient rule so..

    u=2 => du/dx=???? ...do i just put zero?

    It also says write down so it seems to be implying its v easy...i am being daft and missing something..ahhh help my tired brain!!
    t = (3-x^2)
    dt/dx = -2x

    y=2/t = 2t^-1
    dy/dt = -2t^-2 = -2/t^2

    dt/dx x dy/dt = dy/dx
    = 4x/t^2
    =4x/(3-x^2)^2
    x = 1



    4/(3-1)^2 = 1
 
 
 

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