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    (Original post by Ralfskini)
    No, keep trying.
    ok..
    is it

    (x^x^x)(x^(x-1) + (x^x)ln x + (x^x)(ln x)^2) ?
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    (Original post by Ralfskini)
    No, keep trying.
    I get this nasty contraption:

    ( x^x^x ) ln(x^x^x) [ 1 + lnx + 1/(xlnx) ]
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    what about this question from last years summer paper, its easy, but i have cant remember how to do it

    find the cartesian equation and sketch the curve from the following parametric equations

    x = 3cost y=cos2t
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    (Original post by kimoni)
    ok..
    is it

    (x^x^x)(x^(x-1) + (x^x)ln x + (x^x)(ln x)^2) ?

    Yes, that's what I got (although I did make up the question)
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    (Original post by scottyoneill)
    what about this question from last years summer paper, its easy, but i have cant remember how to do it

    find the cartesian equation and sketch the curve from the following parametric equations

    x = 3cost y=cos2t
    y = cos2t = 2cos^2(t)-1 = 2(x/3)^2 - 1
    y = (2/9)x^2-1

    which is a parabola going though (0,-1) and (3/sqrt(2),0), (-3/sqrt(2),0).
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    (Original post by kimoni)
    ok..
    is it

    (x^x^x)(x^(x-1) + (x^x)ln x + (x^x)(ln x)^2) ?
    could you show me step by step?
    i've no idea how i can solve it..
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    here's one for all u then:diffentiate y=x^x^x^x^x^x^x^x^x^x^x.......(g oes on forever)
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    (Original post by ralphk)
    could you show me step by step?
    i've no idea how i can solve it..
    y=x^x^x then lny=ln(x^x^x)=(x^x)lnx
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    (Original post by Mrm.)
    Just use integration by parts with u=x and v´=Cos 2x
    ie u´=1 and v = 1/2 Sin 2x

    therefore

    I xCos2x dx = x . 1/2 Sin 2x - I I/2 Sin 2x dx

    = 1/2 x 2 Sin x . Cos x + 1/4 Cos 2x + c

    = 1/2 x 2 Sin x.Cos x + 1/4 (1 - 2 Sin^2 x) +c

    = 1/2 x 2 Sin x.Cos x + 1/4 - 1/2 Sin^2 x +c

    = 1/2 x 2 Sin x.Cos x - 1/2 Sin^2 x + k

    = 1/2(sinx(2xcosx-sinx) + k
    where did the highlighted cosx come from? :confused:
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    identity: sin 2x = 2 sinx.cosx
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    (Original post by Mrm.)
    identity: sin 2x = 2 sinx.cosx
    oh yea i see... thanks!
 
 
 

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