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    http://www.thestudentroom.co.uk/show....php?t=2705197

    For the S14 IAL F2 Model answers found in the above thread,

    for question 5 part a,

    can someone tell me how arsey goes from 2xy''' to 2x(2xy" )
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    (Original post by Katiee224)
    http://www.thestudentroom.co.uk/show....php?t=2705197

    For the S14 IAL F2 Model answers found in the above thread,

    for question 5 part a,

    can someone tell me how arsey goes from 2xy''' to 2x(2xy" )
    He's differentiating 2xy''' using the product rule.
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    (Original post by Zacken)
    He's differentiating 2xy''' using the product rule.
    but when you differentiate 2xy''' you get 2xy'''' + 2y''' ?
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    (Original post by Katiee224)
    but when you differentiate 2xy''' you get 2xy'''' + 2y''' ?
    Aye, sorry - we have y^{(3)} = 2xy^{(2)} (y^{(3)} means the third derivative of y) differentiating this using the product rule we get: y^{(4)} = 2xy^{(3)} + 2y^{(2)}

    But we know something about y^(3), i.e: that it is the same thing as y^{(3)} = 2xy^{(2)} so we can just replace it into our y^{(4)}, so that:

    \displaystyle

\begin{equation*}y^{(4)} = 2xy^{(3)} + 2y^{(2)} = 2x(2xy^{(2)}) + 2y^{(2)}\end{equation*}
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    (Original post by Zacken)
    Aye, sorry - we have y^{(3)} = 2xy^{(2)} (y^{(3)} means the third derivative of y) differentiating this using the product rule we get: y^{(4)} = 2xy^{(3)} + 2y^{(2)}

    But we know something about y^(3), i.e: that it is the same thing as y^{(3)} = 2xy^{(2)} so we can just replace it into our y^{(4)}, so that:

    \displaystyle

\begin{equation*}y^{(4)} = 2xy^{(3)} + 2y^{(2)} = 2x(2xy^{(2)}) + 2y^{(2)}\end{equation*}
    At the risk of sounding stupid here... why does y^{(3)} = 2xy^{(2)} ?
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    (Original post by Katiee224)
    At the risk of sounding stupid here... why does y^{(3)} = 2xy^{(2)} ?
    The question gives us y'' -2xy'  + 2y = 0 differentiating this:

    \displaystyle

 \begin{equation*}y^{(3)} - 2xy''  - 2y' + 2y' = 0 \iff y^{(3)} = 2xy''\end{equation*}

    by using the product rule on the middle term and re-arranging.
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    (Original post by Zacken)
    The question gives us y'' -2xy'  + 2y = 0 differentiating this:

    \displaystyle

 \begin{equation*}y^{(3)} - 2xy''  - 2y' + 2y' = 0 \iff y^{(3)} = 2xy''\end{equation*}

    by using the product rule on the middle term and re-arranging.
    ohhh they simply got that from the first part of the working and substituted it in to the second

    haha i feel super dumb for not spotting that, appreciate the help
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    (Original post by Katiee224)
    ohhh they simply got that from the first part of the working and substituted it in to the second

    haha i feel super dumb for not spotting that, appreciate the help
    No problem.
 
 
 
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