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Use the first principles definition to determine the first derivative of sqrt(2x-1)

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This is as far as I got:|dw:1340139301477:dw|
Let's look at the definition: \[\lim_{h \rightarrow a} f(x)=[f(x+h) -f(x) ]/h\]

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Other answers:

Your f(x) = sqrt(2x-1), and a=0. So you want to find the limit of f(x) as h approaches 0
How do I know a=0?
So you have \[\lim_{h \rightarrow 0} \sqrt(2x-1)= \sqrt (2(x+h)-1) - \sqrt (2x-1)/h\] where h is the denominator and everything else is on top, the numerator
As for your latter instructions. I understand, and have reached past that point. The reaffirmation is appreciated.
Look at the third picture here: and the explanation below it states why a=0
Ok. I think I understand; The IROC is 0 at a right. Let's continue.
Where are you at @IsTim ? \[\lim_{h \rightarrow 0}\frac{\sqrt{2(x+h)-1}-\sqrt{2x-1}}{h}\] This is where you at right?
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.
Think there is something wrong ... because you should be able to get that h canceled
\[\lim_{h \rightarrow 0}\frac{\sqrt{2x+2h-1}-\sqrt{2x-1}}{h} \cdot \frac{\sqrt{2x+2h-1}+\sqrt{2x-1}}{\sqrt{2x+2h-1}+\sqrt{2x-1}}\]
So that is what you did?
Yes, that was my previous step before this.
\[\lim_{h \rightarrow 0} \frac{(2x+2h-1)-(2x-1)}{h(\sqrt{2x+2h-1}+\sqrt{2x-1})}\] Is this what you got next?
I then cancelled out 2a and -2a.
what else cancels?
on top!
I forgot to cancel 1 and -1.
yep! :)
So that means we have \[\lim_{h \rightarrow 0}\frac{2h}{h(\sqrt{2x+2h-1}+\sqrt{2x-1})}\] Correct?
Do you know what to do from here?
Yes. thank you very much.

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