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    Having trouble with 2. b) in getting the last mark.

    http://store.aqa.org.uk/qual/gce/pdf...W-QP-JUN11.PDF

    http://store.aqa.org.uk/qual/gce/pdf...W-MS-JUN11.PDF

    Cheers.
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     \displaystyle \frac{6cos3x + 6}{(1+cos3x)^2} = \frac{6(cos3x + 1)}{(1+cos3x)^2} = \frac6{1+cos3x}

    Was this the confusion?
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    (Original post by raheem94)
     \displaystyle \frac{6cos3x + 6}{(1+cos3x)^2} = \frac{6(cos3x + 1)}{(1+cos3x)^2} = \frac6{1+cos3x}

    Was this the confusion?
    Thanks, obvious!
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    Do you mean from:

    6cos3x+6/(1+cos3x)^2

    to

    6/1+cos3x



    Because if that's the last mark then 6cos3x+6 can be re-written as 6+6k (let's say cos3x is k).

    so 6+6k/(1+k)^2 which is equal to:

    6(1+k)/(1+k)(1+k)

    take away a (1+k) from top and bottom and you are left with 6/1+k which is:

    6/1+cos3x




    Somehow, if you made it that far, I doubt that is the step you were stuck at so apologies if I just explained something basic.
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    Can anyone help me with 7b
    http://store.aqa.org.uk/qual/gceasa/...W-MS-JAN10.PDF
    http://store.aqa.org.uk/qual/gce/pdf...W-QP-JAN10.PDF

    cheers
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    Try to rearrange the dy/dx into p(1+(tan4x)(tan4x))
    This can then be used in the product rule (or chain, i can't remember which is which.)

    Once you find that rearrange and remember y=tan4x

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