Question

Could someone explain to me the following reasoning about derivatives? (more details in the reply) Could someone explain to me the following reasoning about derivatives? (more details in the reply) @Mathematics

Answer 1

It comes from the following expression:
\[\frac{d}{dx}(\ln |u|)=\frac{1}{u}\frac{du}{dx}\]
From the formula above we obtain the following formula by indefinite integration:
\[\int\limits \frac{1}{u}du=\ln |u|+C\] (C is a constant)
How is the result above obtained from the first expression?

Answer 2

I think I got it now, could someone confirm?
\[d(\ln |u|)=\frac{1}{u}\frac{du}{dx}dx\]
Using the following relationship: \[du=u'(x)dx=\frac{du}{dx}dx\]
\[d(\ln |u|)=\frac{1}{u}du\]
Integrating both sides:
\[\int d(\ln |u|)=\int\frac{1}{u}du\]
\[\ln |u|+C=\int\frac{1}{u}du\]
which is the desired result.
Is this correct?

Answer 3

looks fine to me

Answer 4

Just one question: why are absolute value signs needed? Couldn't I write ln u instead of ln |u|?

Answer 5

In other words, could it be like below?
\[\int \frac{1}{u}du=\ln u + C\]

Answer 6

the absolutes are necessary because u must be positive - the log of a negative does not exist

Answer 7

But, if it is true that
\[\frac{d(\ln u)}{du}=\frac{1}{u}\]
then, why can't I make the reasoning below?
\[d(\ln u) = \frac{1}{u}du\]
\[\int d(\ln u) = \int\frac{1}{u}du\]
\[\ln u+C = \int\frac{1}{u}du\]
what is there in this manipulation that doesn't follow?

Answer 8

let x>0
\[\frac{d}{dx}\ln(x)=\frac{1}{x}\]

let x<0
\[\frac{d}{dx}\ln(-x)=\frac{1}{-x}(-1)=\frac{1}{x}\]

thus \[\frac{d}{dx}\ln|x|=\frac{1}{x}\]

Answer 9

But what is wrong with the manipulation I did above?

Answer 10

for what you have above that is only true if u>0

Answer 11

unless you want your constant, C, to be complex

Answer 12

OK, I think I got it now.