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Derivative with natural log

  1. Jan 11, 2005 #1
    Hi everyone!

    (d/dx) [xlnx]

    Why is the answer 1+lnx and not 1??

    ~Thanks
     
  2. jcsd
  3. Jan 11, 2005 #2

    Curious3141

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    Do you know the product rule ?
     
  4. Jan 11, 2005 #3
    [tex]\frac{d\ln x}{dx} = \frac{1}{x}[/tex], in case you didn't know.
     
  5. Jan 11, 2005 #4
    [tex] f'(x) = x * (1/x) + ln x * 1 = 1 + ln x [/tex]
    [tex] d(uv) = vdu + udv [/tex]
     
  6. Jan 11, 2005 #5
    Wow I'm really stupid!!! I don't know how I did not see that. Thanks :smile:
     
  7. Jan 11, 2005 #6
    At first I did:
    x(1/x) which gave me 1 but why does this not work:

    xlnx
    ln (x^x)
    [x(x^x-1)]/(x^x) ?????

    I know the second way is not how you would usually do it but why does it not work??
     
  8. Jan 11, 2005 #7
    What you've just written is very confusing, do you mind trying to make it a bit more clear?
     
  9. Jan 11, 2005 #8
  10. Jan 11, 2005 #9

    Curious3141

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    [tex]\frac{d}{dx}(x^n) = nx^{n-1}[/tex] only works when n is a constant. Can you see your error now ?
     
  11. Jan 11, 2005 #10
    the derivative of x^x is not [tex] x( x^x - 1)[/tex]. It's [tex] x^x((ln(x) + 1)) [/tex]
     
    Last edited: Jan 11, 2005
  12. Jan 11, 2005 #11
    Yeah I see my error. Thanks.
     
  13. Jan 11, 2005 #12
    Is there a rule for this or something?
     
  14. Jan 11, 2005 #13
    [tex] x^x(ln(x) +1 ) [/tex]

    [tex] f(x) = a^x [/tex] then [tex] f'(x) = a^x(ln(a)) [/tex]
     
  15. Jan 11, 2005 #14
    Thanks :smile:
     
  16. Jan 11, 2005 #15

    Curious3141

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    You don't need a rule.

    [tex]x^x = e^{x\ln x}[/tex]. Can you see how to differentiate it now ?

    The other less direct but "easier to see" way is to use implicit differentiation.

    I once "just used" the derivative of [itex]x^x[/itex] in exam and got docked a couple of points spoiling an otherwise perfect score. The teacher refused to believe I just did it in my head. :tongue2:
     
  17. Jan 11, 2005 #16
    I'm sorry but you just lost me. How did you get [tex]x^x = e^{x\ln x}[/tex]?
     
  18. Jan 11, 2005 #17

    Curious3141

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

    That's actually the easy part. :biggrin:
     
  19. Jan 11, 2005 #18
    :confused: Do you have to memorize this or something?
     
  20. Jan 11, 2005 #19

    Curious3141

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    No, isn't it obvious ? I'm just using [itex]a = e^{\ln a}[/itex] and [itex]{(a^b)}^c = a^{bc}[/itex]
     
  21. Jan 11, 2005 #20
    [tex]x^x = (e^{\ln x})^x = e^{x\ln x}[/tex] there is also a rule which states this. But as Curious said, its pretty obvious.
     
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