Help! derivative of (ln e)

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  • #1
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Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
pwsnafu
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ln y = ln e^(x^x)

(y'/y) = [ (x^x) . (ln e) ]
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?
 
  • #3
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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?
 
  • #4
pwsnafu
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derivative of x^x = x^x ln x
No try again.

(y'/y) = [ (x^x) . (ln e) ]
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.
 
  • #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))'.
 
  • #6
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ok so d/dx (ln e) = 1 not "0"?

if it is 1 then d/dx 1 = 0 ; where d/dx c = 0
 
  • #7
pwsnafu
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ok so d/dx (ln e) = 1 not "0"?
What? How did you get that?

if it is 1 then d/dx 1 = 0 ; where d/dx c = 0 ?
Look ##\frac{d}{dx} \ln e = \frac{d}{dx} 1 = 0##.
The derivative of a constant, any constant, is zero.
 
  • #8
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ok great tnhks
 
  • #9
SteamKing
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Everyone knows that the derivative of ln e is 1/e. :biggrin:
 
  • #10
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Everyone knows that the derivative of ln e is 1/e. :biggrin:
And as a check, ##\int \frac 1 e = \ln e ##:tongue:
 
  • #11
HallsofIvy
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ok so d/dx (ln e) = 1 not "0"?

if it is 1 then d/dx 1 = 0 ; where d/dx c = 0
ln(e) is equal to 1, not the derivative. Because ln(2)= 1, a constant, its derivative is 0.
 

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