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Derivative x^x

  1. Jul 17, 2003 #1
    How do you solve the derivative of x^x? I'm sure it's fairly easy-- I'm just beginning calc though and none of the "forumlas" work.
     
  2. jcsd
  3. Jul 17, 2003 #2
    Take the natural log of both sides.
    y=xx
    lny=xlnx
    Differentiate with respect to x.
    (1/y)(dy/dx)=lnx+x/x (chain rule, product rule)
    dy/dx=y(lnx+1)

    You could do the same thing with y=xa, where a is a constant.
    lny=alnx
    dy/dx=y(a/x)=xa(a/x)=axa-1
    a formula I'm sure you know well
     
    Last edited: Jul 17, 2003
  4. Jul 18, 2003 #3
    or xx = ex*ln(x) ad continue from here...
    (ex*ln(x))' = ex*ln(x) * (x*ln(x))' and so on...
     
  5. Jul 22, 2003 #4

    Dx

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    check out the power rule, learn it , know it, LOVE IT! Its easy, i am just playing but check it out, ok. Any problems with it lemme know, k.
    Dx :wink:
     
  6. Jul 22, 2003 #5

    Tom Mattson

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    Re: Re: derivative x^x



    No, the power rule is not applicable here because the exponent is not a constant. Stephen Privitera's solution is correct; go with that.
     
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