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

]]>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.:P:/

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...

]]>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.

]]>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:

http://www.calculus-help.com/answers/2006_0131.gif

Do note that you shouldn't leave

yin your answer; instead substitutey= 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.

]]>http://www.calculus-help.com/answers/2006_0131.gif

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

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

Anyone want a shot at it?

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Anyway, do you think I got the poblem right?

]]>y = x

Take the derivative of both sides:

y' = x'

But x' = 1 (dx = 1)

y' = 1

]]>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.

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