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The derivative of x^x^x

  1. Mar 13, 2009 #1
    let y=x^x^x
    put natural logarithm ln on both sides
    ln y=ln x^x^x
    we know that ln a^b= b*(ln a)
    let x^x=b
    therefore: ln y= (x^x)*(ln x)
    now,taking the derivative of both sides
    derivative of ln y=y`/y
    and x^x=u,ln x= v.........using the u*v rule we get=u`v+v`u(where ` is the derivative)
    so we have:

    y`/y=(x^x)`* (ln x)+(ln x)` * (x^x) [the derivative of x^x is =x^x(ln x+1) and derivative of ln x=1/x]

    so, y`/y= x^x(ln x+1)*(ln x)+(1/x)*(x^x)

    y`/y=x^x(ln^2 x+ln x)+(1/x)*(x^x)
    taking x^x common on the right hand side
    y`/y=x^x[ln^2 x +ln x +1/x]
    now the y dividing y` goes to the right hand side and multiplies
    y`=y*(x^x)[ln^2 x +ln x+1/x]
    we know that y=x^x^x
    therefore y`=(x^x^x)*(x^x)[ln^2 x +ln x+1/x]
    since the bases are same ,powers should be added
    so the final answer becomes
    y`=x^[(x^x)+x][ln^2 x +ln x+1/x]
    OR
    d(y)/dx=x^[(x^x)+x][ln^2 x +ln x+1/x]
     
  2. jcsd
  3. Mar 13, 2009 #2

    HallsofIvy

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    I assume you mean x^(x^x) since (x^x)^x= x^(x^2)

    Looks good to me.
     
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