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    a) Defferenciate x^(x^2) with respect to x

    is this **** done in further maths
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    (Original post by matouwah)
    a) Defferenciate x^(x^2) with respect to x

    is this **** done in further maths

    what module is it?
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    (Original post by matouwah)
    a) Defferenciate x^(x^2) with respect to x

    is this **** done in further maths
    let y=x^2

    (1/y)y'=2xlnx + x (d/dx on both sides)

    => y'= x^(x^2)[2xlnx+x]
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    (Original post by ShOcKzZ)
    what module is it?
    dunno....some guy doing the course i want to do at oxford was asked it during the interview....but he was doing further maths.....i havent ecountered this type of question on p3....and i'm not doing further maths...
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    (Original post by M Safe)
    let y=x^2

    (1/y)y'=2xlnx + x (d/dx on both sides)

    => y'= x^(x^2)[2xlnx+x]
    sorry y=x^(x^2)
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    (Original post by M Safe)
    let y=x^2

    (1/y)y'=2xlnx + x (d/dx on both sides)

    => y'= x^(x^2)[2xlnx+x]
    uhmmm.....kinda lost....could u explain it in more detail...i am a bit of a retard.....thanx...
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    (Original post by matouwah)
    uhmmm.....kinda lost....could u explain it in more detail...i am a bit of a retard.....thanx...
    y=x^(x^2)

    ln y = (x^2)ln(x)

    differentiate wrt x on both sides (dy/dx = d/dy * dy/dx)

    so 1/y * (dy/dx) = x + 2xln(x) (use product rule to obtain RHS)

    do dy/dx = y(x+2xln(x))
    = ...
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    (Original post by M Safe)
    y=x^(x^2)

    ln y = (x^2)ln(x)

    differentiate wrt x on both sides (dy/dx = d/dy * dy/dx)

    so 1/y * (dy/dx) = x + 2xln(x) (use product rule to obtain RHS)

    do dy/dx = y(x+2xln(x))
    = ...

    ahhhhh i see it now.....thanx...
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    take logs.
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    i once differentiated x^x^x "for fun". woohoo the hilarity! (i was bored in a free period). i then went on to do x^x^x^x, and their integrals. god i am sad.
 
 
 
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