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Differentiating y=x^x

  1. Aug 15, 2012 #1
    Differentiating y=x^x

    x=ln(y)

    I changed the base to e

    [tex]x=\frac{ln(y)}{ln(x)}[/tex]
    [tex]xln(x) = ln(y)[/tex]
    [tex]e^{xln(x)} = y[/tex]
    [tex]e^{xln(x)}(1+ln(x) = \frac{dy}{dx}[/tex]

    The answer the calculator got was [tex]x^{x(1+ln(x))}[/tex] so I noticed that since [tex]y=x^x[/tex] and [tex]e^{xln(x)} = y[/tex], then I could replace it with x^x in the final answer

    Is this an acceptable method? Is there any circular logic I missed? Could I leave it as is wihtout writing x^x?
     
  2. jcsd
  3. Aug 15, 2012 #2
    yes, but the factor [itex]1+\ln x[/itex] should not be in the exponent (chain rule).
     
  4. Aug 15, 2012 #3
    ^ Thank you...yeah I made an error there
     
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