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## #1 2006-01-12 14:21:11

mikau
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### logarithmic differentiation

Differentiate y = 2x^(2x) using logarithmic differentiation.

I've done it several times, in different ways, but I can't seem to get the correct answer.

A logarithm is just a misspelled algorithm.

## #2 2006-01-12 15:03:56

mikau
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### Re: logarithmic differentiation

I keep getting dy/dx = y2(1 + ln 2x)

My book gives an answer of y2(1 + ln x)

A logarithm is just a misspelled algorithm.

## #3 2006-01-12 15:19:37

Ricky
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### Re: logarithmic differentiation

The only way I know how to solve this is implicit differentiation:

Let y = x^x.

ln(y) = ln(x^x)

ln(y) = x*ln(x)

The derivative of ln(y) is 1/y, but since it's implicit, we need to use the chain rule on y, which means we multiply it by the derivative of y.  It then becomes 1/y * y'

The derivative of x*ln(x) using the product rule is x*1/x + ln(x) * 1.  This is the same as 1 + ln(x)

So we have:

y' / y = 1 + ln(x)

But we know that y = x^x

y' / x^x = 1 + ln(x)

y' = (1 + ln(x)) * x^x

Now just go through the same steps using 2's.

Edit: To check your answer, the answer to 2(x)^(2x) is 4(ln(x)+1)x^(2x) and the answer to (2x)^(2x) is (2ln(2x)+2)*(2x)^(2x), I'm not sure which one you meant.

Last edited by Ricky (2006-01-12 15:26:50)

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

## #4 2006-01-12 16:06:21

mikau
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### Re: logarithmic differentiation

"The only way I know how to solve this is implicit differentiation:"

Yeah but you took the natural log of both sides to simplify, so you did use logarithmic differentiation.

But man, you did something wierd. You took the derivative of ln y, and wrote 1/y * y' in accordance with the chain rule. I suppose this is the same as 1/y dy but you didn't do it on the other side. (dx)

Anyways, I'll attempt to do it using your method.

u = 2x

y = u^u

ln y = u ln u

1/y dy = u/u du + ln u du

1/y dy = du + ln u du
1/y dy = (1 + ln u) du

Now  u = 2x so du = 2 dx

Inserting these values we get:

1/y dy =  (1 + ln 2x)2 dx

dy/dx = y(1 + ln 2x) 2

Once again, I arrive at the same conclusion.

A logarithm is just a misspelled algorithm.

## #5 2006-01-13 06:45:56

mikau
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### Re: logarithmic differentiation

So you think maybe its a misprint?

A logarithm is just a misspelled algorithm.

## #6 2006-01-13 12:03:02

Ricky
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### Re: logarithmic differentiation

Since your taking the derivative with respect to x, dx = 1.  But you aren't taking the derivative with respect to y, so dy doesn't have to be 1.  For example:

y = x

Take the derivative of both sides:

y' = x'

But x' = 1 (dx = 1)

y' = 1

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

## #7 2006-01-13 12:18:49

mikau
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### Re: logarithmic differentiation

Seems to arrive at the same place. But I'm not familar with that method.

Anyway, do you think I got the poblem right?

A logarithm is just a misspelled algorithm.

## #8 2006-01-14 15:02:57

John E. Franklin
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### Re: logarithmic differentiation

I found this but don't know if it is on the same topic, nor do I know how to derive it.

Imagine for a moment that even an earthworm may possess a love of self and a love of others.

## #9 2006-01-16 08:48:57

mikau
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### Re: logarithmic differentiation

In case no one noticed, this still has not been solved. I'm pretty confident I'm doing it right but my book always end up right in the end.

Differentiate y = 2x^(2x) using logarithmic differentiation.

Anyone want a shot at it?

A logarithm is just a misspelled algorithm.

## #10 2006-02-01 05:00:07

Mike
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### Re: logarithmic differentiation

Hey there, just stumbled across your forum. I run the web site www.calculus-help.com and am happy to post a solution. Just follow this link:

Do note that you shouldn't leave y in your answer; instead substitute y = 2x^(2x) back in for it at the end

## #11 2006-02-01 05:02:48

Calculus-Help
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### Re: logarithmic differentiation

#### Mike wrote:

Hey there, just stumbled across your forum. I run the web site www.calculus-help.com and am happy to post a solution. Just follow this link:

Do note that you shouldn't leave y in your answer; instead substitute y = 2x^(2x) back in for it at the end

I just went ahead and registered after I posted the above solution, so you can direct comments and questions to me at this username.

## #12 2006-02-01 06:32:51

mikau
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### Re: logarithmic differentiation

Many thanks, Mike. That is the same answer I got. But the answer the book gives is:

4x^(2x) (1 + ln x)   which is exactly the same except it says ln x instead of ln (2x).

At this point I'd say it could be a misprint but I've said that before and my book always ends up being write in the end. Still, it could happen.

A logarithm is just a misspelled algorithm.

## #13 2006-02-01 07:50:30

Calculus-Help
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### Re: logarithmic differentiation

I have to concur -- it should be a 2x in there. Having published a handful of math books, I must say it is not unreasonable to assume it is a misprint. They happen relatively often in math publishing, because the vast majority of book proofreaders don't understand math at all, which is kind of sad.

## #14 2006-02-01 08:25:46

mikau
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### Re: logarithmic differentiation

You've published mathbooks?

Well if this truely is a misprint, it's the first time I've ever seen one in a Saxon book. Like I said, usually I think its a misprint, but in the end it always turns out right. Saxon's been completely reliable up to now.

But it is odd how most other mathbooks are written by highschool dropouts...

A logarithm is just a misspelled algorithm.

Browniegirl32296
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No I havnet.

## #16 2006-02-01 08:37:00

Browniegirl32296
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### Re: logarithmic differentiation

#### mikau wrote:

Differentiate y = 2x^(2x) using logarithmic differentiation.

I've done it several times, in different ways, but I can't seem to get the correct answer.

I don't know the answer either. Sorry :-D
My teacher has not gotten in to that.:/

## #17 2006-02-01 08:38:15

Browniegirl32296
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### Re: logarithmic differentiation

Help me!! I can't find a website that has work out sheets you can print out. Do you people have any?

## #18 2006-02-01 08:44:00

mikau
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### Re: logarithmic differentiation

You could have posted a new thread for this. :-)

Some worksheets here: http://www.mathsisfun.com/forum/viewtopic.php?id=483

A logarithm is just a misspelled algorithm.