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Derivative of x to the x

  1. Feb 12, 2007 #1
    1. The problem statement, all variables and given/known data

    I am required to find the derivative of x^x using the differance quotient

    2. Relevant equations


    3. The attempt at a solution

    TI-89 gives (lnx+1)x^x I can't figure out how to get there
  2. jcsd
  3. Feb 12, 2007 #2
    Without the derivative definition, ln both sides, so:

    [tex]\frac{1}{y}.\frac{dy}{dx} = 1.ln(x) + x.\frac{1}{x}[/tex]
    [tex]\frac{dy}{dx} = y[ln(x) + 1][/tex]
    [tex]\frac{dy}{dx} = x^x(ln(x) + 1) [/tex]
  4. Feb 13, 2007 #3


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    I can't imagine there being any reasonable way to use the "difference quotient" to determine the derivative of xx. You must have a really evil teacher!
  5. Feb 14, 2007 #4


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    Either this is unreasonable, or your imagination really needs to see more:

    [tex] x^x =e^{x\ln x} [/tex]

    [tex] \lim_{h\rightarrow 0} \frac{(x+h)^{(x+h)}-x^x}{h} =\lim_{h\rightarrow 0} \frac{e^{(x+h)\ln(x+h)}-e^{x\ln x}}{h}=\lim_{h\rightarrow 0}\frac{e^{(x+h)\ln\left(x\left(1+\frac{h}{x}\right)\right)}-e^{x\ln x}}{h}[/tex]

    [tex] =\lim_{h\rightarrow 0} \frac{e^{x\ln x}e^{h\ln x}e^{x\ln\left(1+\frac{h}{x}\right)}e^{h\ln\left(1+\frac{h}{x}\right)}-e^{x\ln x}}{h} [/tex]

    [tex] =x^{x}\lim_{h\rightarrow 0}\frac{e^{h\ln x}e^{x\ln\left(1+\frac{h}{x}\right)}e^{h\ln\left(1+\frac{h}{x}\right)}-1}{h}\cdot \frac{h\ln x+x\ln\left(1+\frac{h}{x}\right)+h\ln\left(1+\frac{h}{x}\right)}{h\ln x+x\ln\left(1+\frac{h}{x}\right)+h\ln\left(1+\frac{h}{x}\right)} [/tex]

    [tex] =x^{x}\left[\ln x+x\lim_{h\rightarrow 0}\frac{\ln\left(1+\frac{h}{x}\right)}{h}+\lim_{h\rightarrow 0}\ln\left(1+\frac{h}{x}\right)\right] [/tex]

    [tex] =x^x \left\{\ln x+x\ln \lim_{h\rightarrow 0}\left[\left(1+\frac{h}{x}\right)^{\frac{x}{h}}\right]^{\frac{1}{x}}\right\} [/tex]

    [tex] =x^x \left(\ln x +1\right) [/tex]
    Last edited: Feb 14, 2007
  6. Feb 15, 2007 #5
    wow .
  7. Feb 15, 2007 #6


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    Whoa. :bugeye:
    Or somewhat easier. That's a little bit messy methinks: o:)
    [tex](x ^ x)' = (e ^ {x \ln (x)})' = \lim_{h \rightarrow 0} \frac{e ^ {(x + h) \ln (x + h)} - e ^ {x \ln (x)}}{h}[/tex]

    [tex]= \lim_{h \rightarrow 0} \left( \frac{e ^ {(x + h) \ln (x + h)} - e ^ {x \ln (x)}}{(x + h) \ln (x + h) - x \ln x} \times \frac{(x + h) \ln (x + h) - x \ln x}{h} \right)[/tex]

    [tex]= e ^ {x \ln (x)} \left( \lim_{h \rightarrow 0} \frac{x \ln (1 + \frac{h}{x}) + h \ln (x + h)}{h} \right)[/tex] [due to: (eu)'u = eu]

    [tex]= x ^ x \left\{ \lim_{h \rightarrow 0} \left[ x \ln \left( \left( 1 + \frac{h}{x} \right) ^ {\frac{x}{h}} \right) ^ {\frac{1}{x}} \right] + \ln (x) \right\}[/tex]

    [tex]= x ^ x (1 + \ln (x))[/tex]
    Last edited: Feb 15, 2007
  8. Aug 21, 2009 #7
    Since the original equation is x^x and never a division quotient, one can not apply that rule. Here is the alternative proof.

    Attached Files:

  9. Aug 22, 2009 #8
    "Difference quotient" means this:
    so the assignment was to use the definition of derivative that involves this.
  10. Aug 23, 2009 #9
    An answer is attached in PDF

    Attached Files:

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