The title for this content could be: Understanding the Derivative of ln(e)

[dimasalang]

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

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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) = 1anwers 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.

 

[pwsnafu]

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?

 

[dimasalang]

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?

 

[pwsnafu]

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.

 

[HallsofIvy]

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 $e^{x^x}$". You seem to be trying to use the general "derivative of $a^x$ is $a^x ln(a)$ but it is much simpler, for the special case of a= e, to use "the derivative of $e^x$ is $e^x$. That "special case" is important enough to memorize by itself.

So the derivative of $e^{x^x}$ is, by the chain ruie, $e^{x^x}$ times the derivative of $x^x$. That is the one pwsnafu has been trying to get you to do. If $y= x^x$, then $ln(y)= ln(x^x)= x ln(x)$. 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))'.

 

[dimasalang]

ok so d/dx (ln e) = 1 not "0"?

if it is 1 then d/dx 1 = 0 ; where d/dx c = 0

 

[pwsnafu]

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.

 

[dimasalang]

ok great tnhks

 

[SteamKing]

Everyone knows that the derivative of ln e is 1/e.

 

[Mark44]

Everyone knows that the derivative of ln e is 1/e.

And as a check, $\int \frac 1 e = \ln e$

 

[HallsofIvy]

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.