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IsTim
Group Title
Use the first principles definition to determine the first derivative of sqrt(2x1)
 2 years ago
 2 years ago
IsTim Group Title
Use the first principles definition to determine the first derivative of sqrt(2x1)
 2 years ago
 2 years ago

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IsTim Group TitleBest ResponseYou've already chosen the best response.0
This is as far as I got:dw:1340139301477:dw
 2 years ago

mashe Group TitleBest ResponseYou've already chosen the best response.0
Let's look at the definition: \[\lim_{h \rightarrow a} f(x)=[f(x+h) f(x) ]/h\]
 2 years ago

mashe Group TitleBest ResponseYou've already chosen the best response.0
Your f(x) = sqrt(2x1), and a=0. So you want to find the limit of f(x) as h approaches 0
 2 years ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
How do I know a=0?
 2 years ago

mashe Group TitleBest ResponseYou've already chosen the best response.0
So you have \[\lim_{h \rightarrow 0} \sqrt(2x1)= \sqrt (2(x+h)1)  \sqrt (2x1)/h\] where h is the denominator and everything else is on top, the numerator
 2 years ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
As for your latter instructions. I understand, and have reached past that point. The reaffirmation is appreciated.
 2 years ago

mashe Group TitleBest ResponseYou've already chosen the best response.0
Look at the third picture here: http://www.intmath.com/differentiation/3derivativefirstprinciples.php and the explanation below it states why a=0
 2 years ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
Ok. I think I understand; The IROC is 0 at a right. Let's continue.
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
Where are you at @IsTim ? \[\lim_{h \rightarrow 0}\frac{\sqrt{2(x+h)1}\sqrt{2x1}}{h}\] This is where you at right?
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
You need to rationalize the numerator. We are trying to manipulate this problem so that when we plug in 0 for h we don't get 0 on the bottom.
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
Think there is something wrong ... because you should be able to get that h canceled
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
\[\lim_{h \rightarrow 0}\frac{\sqrt{2x+2h1}\sqrt{2x1}}{h} \cdot \frac{\sqrt{2x+2h1}+\sqrt{2x1}}{\sqrt{2x+2h1}+\sqrt{2x1}}\]
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
So that is what you did?
 2 years ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
Yes, that was my previous step before this.
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
\[\lim_{h \rightarrow 0} \frac{(2x+2h1)(2x1)}{h(\sqrt{2x+2h1}+\sqrt{2x1})}\] Is this what you got next?
 2 years ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
I then cancelled out 2a and 2a.
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
what else cancels?
 2 years ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
I forgot to cancel 1 and 1.
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
So that means we have \[\lim_{h \rightarrow 0}\frac{2h}{h(\sqrt{2x+2h1}+\sqrt{2x1})}\] Correct?
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
Do you know what to do from here?
 2 years ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
Yes. thank you very much.
 2 years ago
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