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lingyai
science forum beginner
Joined: 14 Apr 2006
Posts: 21
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 Posted: Mon May 22, 2006 7:36 am Post subject:
need help checking a basic integration formula via differentiation
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I'm stuck on something simple, I'd be grateful for any help.
My text ("Forgotten Calculus", Bleau) states the following integration formula:
integral [dx / (x ln x)] = ln |ln x| + C.
In the course of a problem where I've applied the formula, I realised I can't prove it.
Here's my attempt, for what it's worth.
(Re notation:
exponents are denoted with the "^" sign, e.g x squared = x^2;
first derivatives are denoted with the " ' " symbol, e.g. the 1st derivative of f(x) = f'(x))
f(x) = ln |ln x| + C
I want to differentiate this via the product rule, so I rewrite the function as
f(x) = (ln x^0) * (ln x) + C
Using the product rule,
f'(x) = [ (ln x^0) * (ln x)' ] + [(ln x^0)' * (ln x) ]
= [ (ln 1) * (1/x)] + [ (0/1) * (ln x)]
= [ ln / x ] + [0]
= ln / x.
I don't see how ln / x = ln |ln x|.
Any advice? |
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Jean-Marc Gulliet
science forum beginner
Joined: 28 May 2005
Posts: 38
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| Posted: Mon May 22, 2006 8:56 am Post subject:
Re: need help checking a basic integration formula via differentiation
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lingyai wrote:
| Quote: | I'm stuck on something simple, I'd be grateful for any help.
My text ("Forgotten Calculus", Bleau) states the following integration formula:
integral [dx / (x ln x)] = ln |ln x| + C.
In the course of a problem where I've applied the formula, I realised I can't prove it.
Here's my attempt, for what it's worth.
(Re notation:
exponents are denoted with the "^" sign, e.g x squared = x^2;
first derivatives are denoted with the " ' " symbol, e.g. the 1st derivative of f(x) = f'(x))
f(x) = ln |ln x| + C
I want to differentiate this via the product rule, so I rewrite the function as
f(x) = (ln x^0) * (ln x) + C
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Why are you using the product rule here? The chain rule seems definitely
more appropriate: f[g[x]]' = f'[g[x]] * g'[x].
Moreover, since the function contains an absolute value and to be
rigorous, you should considered two cases: x > 1 and 0 < x < 1.
That is, if x > 1 we have
Log[x} > 0, therefore f[x] = Log[Log[x]];
and if 0 < x < 1, we have
Log[x] < 0, therefore f[x] = Log[-Log[x]].
Then, applying the chain to both cases should be straightforward: in the
first case, g[x] = Log[x]; and g[x] = -Log[x] in the second case.
HTH,
Jean-Marc |
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David C. Ullrich
science forum Guru
Joined: 28 Apr 2005
Posts: 2250
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| Posted: Mon May 22, 2006 11:09 am Post subject:
Re: need help checking a basic integration formula via differentiation
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On Mon, 22 May 2006 03:36:23 EDT, lingyai <ken_kasriel@yahoo.com>
wrote:
| Quote: | I'm stuck on something simple, I'd be grateful for any help.
My text ("Forgotten Calculus", Bleau) states the following integration formula:
integral [dx / (x ln x)] = ln |ln x| + C.
In the course of a problem where I've applied the formula, I realised I can't prove it.
Here's my attempt, for what it's worth.
(Re notation:
exponents are denoted with the "^" sign, e.g x squared = x^2;
first derivatives are denoted with the " ' " symbol, e.g. the 1st derivative of f(x) = f'(x))
f(x) = ln |ln x| + C
I want to differentiate this via the product rule, so I rewrite the function as
f(x) = (ln x^0) * (ln x) + C
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That "rewriting" is wrong. f(x) is not what the rewrite says, f(x) is
ln(|ln(x)|).
For started restrict to the case x > a, so f(x) = ln(ln(x)).
| Quote: | Using the product rule,
f'(x) = [ (ln x^0) * (ln x)' ] + [(ln x^0)' * (ln x) ]
= [ (ln 1) * (1/x)] + [ (0/1) * (ln x)]
= [ ln / x ] + [0]
= ln / x.
I don't see how ln / x = ln |ln x|.
Any advice?
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************************
David C. Ullrich |
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Badger
science forum beginner
Joined: 07 May 2006
Posts: 38
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| Posted: Mon May 22, 2006 1:19 pm Post subject:
Re: need help checking a basic integration formula via differentiation
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On Mon, 22 May 2006 03:36:23 EDT, lingyai <ken_kasriel@yahoo.com>
wrote:
| Quote: | I'm stuck on something simple, I'd be grateful for any help.
My text ("Forgotten Calculus", Bleau) states the following integration formula:
integral [dx / (x ln x)] = ln |ln x| + C.
In the course of a problem where I've applied the formula, I realised I can't prove it.
Here's my attempt, for what it's worth.
(Re notation:
exponents are denoted with the "^" sign, e.g x squared = x^2;
first derivatives are denoted with the " ' " symbol, e.g. the 1st derivative of f(x) = f'(x))
f(x) = ln |ln x| + C
I want to differentiate this via the product rule, so I rewrite the function as
f(x) = (ln x^0) * (ln x) + C
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Others have already pointed out that this 'rewrite' is not correct.
| Quote: |
Using the product rule,
f'(x) = [ (ln x^0) * (ln x)' ] + [(ln x^0)' * (ln x) ]
= [ (ln 1) * (1/x)] + [ (0/1) * (ln x)]
= [ ln / x ] + [0]
= ln / x.
I don't see how ln / x = ln |ln x|.
Any advice?
|
Others have also pointed out that all you need to do is apply the
chain rule. As a learning tool, you might want to check out the
following site:
<http://www.calc101.com/>
For the question you posted, click on 'derivatives' and then enter
your function, written as
ln[ ln[x] ]
The step-by-step result should make clear to you how the chain rule
is used to find the derivative. |
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lingyai
science forum beginner
Joined: 14 Apr 2006
Posts: 21
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| Posted: Mon May 22, 2006 2:21 pm Post subject:
Re: need help checking a basic integration formula via differentiation
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Thanks everyone for helping me out. And to you, Badger, in particular -- that is a seriously amazing website. Thanks so much for the steer!
Ken |
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Stan Brown
science forum Guru Wannabe
Joined: 06 May 2005
Posts: 279
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| Posted: Tue May 23, 2006 10:34 pm Post subject:
Re: need help checking a basic integration formula via differentiation
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Mon, 22 May 2006 10:56:14 +0200 from Jean-Marc Gulliet
<jeanmarc.gulliet@gmail.com>:
| Quote: | lingyai wrote:
f(x) = ln |ln x| + C
I want to differentiate this via the product rule,
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No, you don't.
ln |ln x| is *not* a product, because "ln" is not a quantity.
ln |ln x| is a composite function, the natural logarithm of |ln x|.
You can rewrite it as
f(x) = g( h(x) )
where g(u) = ln(u) and h(x) = |ln x|.
This is a *very* common category error by students -- it leads to the
sort of nonsense like "canceling" sin in
sin(2x)
-------
sin(x)
to get the answer of "2".
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com/ |
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