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Use the first principles definition to determine the first derivative of sqrt(2x1)
 one year ago
 one year ago
Use the first principles definition to determine the first derivative of sqrt(2x1)
 one year ago
 one year ago

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IsTimBest ResponseYou've already chosen the best response.0
This is as far as I got:dw:1340139301477:dw
 one year ago

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

masheBest 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
 one year ago

masheBest 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
 one year ago

IsTimBest 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.
 one year ago

masheBest 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
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
Ok. I think I understand; The IROC is 0 at a right. Let's continue.
 one year ago

myininayaBest 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?
 one year ago

myininayaBest 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.
 one year ago

myininayaBest ResponseYou've already chosen the best response.3
Think there is something wrong ... because you should be able to get that h canceled
 one year ago

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

myininayaBest ResponseYou've already chosen the best response.3
So that is what you did?
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
Yes, that was my previous step before this.
 one year ago

myininayaBest 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?
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
I then cancelled out 2a and 2a.
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
I forgot to cancel 1 and 1.
 one year ago

myininayaBest 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?
 one year ago

myininayaBest ResponseYou've already chosen the best response.3
Do you know what to do from here?
 one year ago

IsTimBest ResponseYou've already chosen the best response.0
Yes. thank you very much.
 one year ago
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