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Help! derivative of (ln e)

  1. Oct 4, 2013 #1
    please enlighten me, what is the derivative of (ln e) ; natural log e

    is it 0 or 1?

    i have this problem in the book

    y = e^(x^x)

    and the ANSWER is

    y' = (x^x) e^(x^x) [1 + lnx]


    my FIRST solution and i assumed (ln e=1) ; i treated it a constant so it must be "0"

    where (d/dx) c = 0
    b^x = r ---> x = logb r
    (d/dx) ln x = (1/x) (du/dx)

    ln y = ln e^(x^x) <--- i use ln both sides

    (y'/y) = [ (x^x) . (ln e) ] <--- product rule

    u v ------> du v + u dv

    (y'/y) = [(x^x ln x) (ln e)] + [(x^x) (1/e) (e.0)]<---must be zero ;(d/dx) ln e = 0 because ln e = 1


    (y'/y) = [ (x^x ln x) (ln e) + 0 ]


    y' = [(x^x ln x) (1)] (y) ---> multiply by (y/y) to the to both side to get y'

    my final answerr is;

    y' = (x^x ln x) [ e^(x^x) ] <------ where [y = e^(x^x)]

    = which is in contradiction to the ANSWER IN THE BOOK

    =============================================================================
    and the other solution i came up with, to get the right answer in the book is;

    IF (d/dx) (ln e) is = to 1

    ln y = ln e^(x^x)

    (y'/y) = [ (x^x) . (ln e) ]

    (y'/y) = [(x^x ln x) (ln e)] + [(x^x) (1)]

    (y'/y) = [(x^x ln x) (ln e)] + [(x^x)]

    y' = [(x^x ln x) (1)] + [(x^x)] (y) ;(ln e) = 1


    anwers is


    y' = (x^x) e^(x^x) [1 + lnx] ;factored out (x^x) and y = e^(x^x)

    which conforms to the answer in the book







    is my solution correct? pls enlighten me.
     
  2. jcsd
  3. Oct 4, 2013 #2

    pwsnafu

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    You took the derivative of the left, but you haven't taken the derivative on the right.

    P.S. Do you know what the derivative of ##x^x## is equal to?
     
  4. Oct 4, 2013 #3
    derivative of x^x = x^x ln x

    (y'/y) = [ (x^x) . (ln e) ]

    (y'/y) = [(x^x ln x) (ln e)] + [(x^x) (1/e) (e.0)] <-- is this right d/dx of (ln e) is 0?
     
  5. Oct 4, 2013 #4

    pwsnafu

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    No try again.

    Firstly, you haven't taken the derivative on the right like I asked you.

    Serious question: why are writing ln e when clearly you know that it is equal to 1?

    P.S. Also you know what the derivative of 1 is equal to right? So obviously ln e = 1 so the derivative of ln e is equal to the derivative of 1.
     
  6. Oct 4, 2013 #5

    HallsofIvy

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    Your original question "what is the derivative of ln(e)" is easy: ln(e)= 1 is a number, a constant. And the derivative of any constant is 0.

    Of course, that has nothing to do with the "derivative of [itex]e^{x^x}[/itex]". You seem to be trying to use the general "derivative of [itex]a^x[/itex] is [itex]a^x ln(a)[/itex] but it is much simpler, for the special case of a= e, to use "the derivative of [itex]e^x[/itex] is [itex]e^x[/itex]. That "special case" is important enough to memorize by itself.

    So the derivative of [itex]e^{x^x}[/itex] is, by the chain ruie, [itex]e^{x^x}[/itex] times the derivative of [itex]x^x[/itex]. That is the one pwsnafu has been trying to get you to do. If [itex]y= x^x[/itex], then [itex]ln(y)= ln(x^x)= x ln(x)[/itex]. Differentiating on the left, we have y'/y as you say. To differentiate on the right, use the product rule: (x ln(x))'= (x)' ln(x)+ x(ln(x))'.
     
  7. Oct 4, 2013 #6
    ok so d/dx (ln e) = 1 not "0"?

    if it is 1 then d/dx 1 = 0 ; where d/dx c = 0
     
  8. Oct 4, 2013 #7

    pwsnafu

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    What? How did you get that?

    Look ##\frac{d}{dx} \ln e = \frac{d}{dx} 1 = 0##.
    The derivative of a constant, any constant, is zero.
     
  9. Oct 4, 2013 #8
    ok great tnhks
     
  10. Oct 4, 2013 #9

    SteamKing

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    Everyone knows that the derivative of ln e is 1/e. :biggrin:
     
  11. Oct 4, 2013 #10

    Mark44

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    And as a check, ##\int \frac 1 e = \ln e ##:tongue:
     
  12. Oct 4, 2013 #11

    HallsofIvy

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    ln(e) is equal to 1, not the derivative. Because ln(2)= 1, a constant, its derivative is 0.
     
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