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Differentiate x^(x^2)

  1. Jun 8, 2009 #1
    1. The problem statement, all variables and given/known data
    Differentiate [tex]x^{x^2}[/tex], with respect to x


    2. Relevant equations

    [tex]\frac{d}{dx} (x^{x^2})[/tex]


    3. The attempt at a solution
    I arrived at... (ready?)

    (ta dah!): [tex]x^{x^2}.(2x.\ln(x)+x)[/tex]

    I'm pretty confident this is wrong...

    I went [tex]y(x)=x^{x^2}[/tex], then took the natural logarithm of both sides

    Since [tex]\ln(x^{x^2}) = x^2.\ln(x) and \frac{d}{d(y(x))} (ln (y(x))) = \frac{1}{y(x)}[/tex]

    I got:

    [tex]\frac{1}{y(x)} . \frac{d(y(x))}{dx} = \frac{d}{dx} (x^2.\ln(x))[/tex]

    [tex]\frac{1}{y(x)} . \frac{d(y(x))}{dx} = 2x.\ln(x)+\frac{x^2}{x}[/tex]

    [tex]\frac{d(y(x))}{dx} = 2x.\ln(x)+\frac{x^2}{x} . x^{x^2} = x^{x^2}.(2x.\ln(x)+x)[/tex]

    As I said, I think this is wrong. I've been working through examples all day and figured I might be able to come back to it, and hopefully figure it out (that is my excuse, and I'm sticking with it! :smile:); but since I don't have easy access to a computer, I thought I might ask you guys now, and check back tomorrow to see if anyone has offered any help.
    Thanks everyone.
     
  2. jcsd
  3. Jun 8, 2009 #2
    Eh I think that is right, though I did it in my head. It's easy if you write it as the exponential of (x^2)(ln x), from which you get the original expression times the derivative of (x^2)(ln x) which by the product rule is 2x(ln x) + x.
     
  4. Jun 8, 2009 #3

    Cyosis

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    I don't see why you think this would be wrong, because it's not.
     
  5. Jun 8, 2009 #4
    Yup, I think it's pretty correct.
     
  6. Jun 9, 2009 #5
    Snipez90, Cyosis & Karkas: Thankyou.
    The question was from a multiple choice "quiz" that I saw, and I just didn't recognise the answer that I got (above) from being one of the possibilities: that's why I thought it was wrong. But it seems its not! Thanks for the responses everyone!
     
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