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IsTim

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

  • one year ago
  • one year ago

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  1. IsTim
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    This is as far as I got:|dw:1340139301477:dw|

    • one year ago
  2. mashe
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    Let's look at the definition: \[\lim_{h \rightarrow a} f(x)=[f(x+h) -f(x) ]/h\]

    • one year ago
  3. IsTim
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    Ok.

    • one year ago
  4. mashe
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    Your f(x) = sqrt(2x-1), and a=0. So you want to find the limit of f(x) as h approaches 0

    • one year ago
  5. IsTim
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    How do I know a=0?

    • one year ago
  6. mashe
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    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

    • one year ago
  7. IsTim
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    As for your latter instructions. I understand, and have reached past that point. The reaffirmation is appreciated.

    • one year ago
  8. mashe
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    Look at the third picture here: http://www.intmath.com/differentiation/3-derivative-first-principles.php and the explanation below it states why a=0

    • one year ago
  9. IsTim
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    Ok. I think I understand; The IROC is 0 at a right. Let's continue.

    • one year ago
  10. IsTim
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    Hello?

    • one year ago
  11. myininaya
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    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?

    • one year ago
  12. myininaya
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    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
  13. myininaya
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    Think there is something wrong ... because you should be able to get that h canceled

    • one year ago
  14. myininaya
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    \[\lim_{h \rightarrow 0}\frac{\sqrt{2x+2h-1}-\sqrt{2x-1}}{h} \cdot \frac{\sqrt{2x+2h-1}+\sqrt{2x-1}}{\sqrt{2x+2h-1}+\sqrt{2x-1}}\]

    • one year ago
  15. myininaya
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    So that is what you did?

    • one year ago
  16. IsTim
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    Yes, that was my previous step before this.

    • one year ago
  17. myininaya
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    \[\lim_{h \rightarrow 0} \frac{(2x+2h-1)-(2x-1)}{h(\sqrt{2x+2h-1}+\sqrt{2x-1})}\] Is this what you got next?

    • one year ago
  18. IsTim
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    I then cancelled out 2a and -2a.

    • one year ago
  19. myininaya
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    what else cancels?

    • one year ago
  20. myininaya
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    on top!

    • one year ago
  21. IsTim
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    I forgot to cancel 1 and -1.

    • one year ago
  22. myininaya
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    yep! :)

    • one year ago
  23. myininaya
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    So that means we have \[\lim_{h \rightarrow 0}\frac{2h}{h(\sqrt{2x+2h-1}+\sqrt{2x-1})}\] Correct?

    • one year ago
  24. IsTim
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    yes.

    • one year ago
  25. myininaya
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    Do you know what to do from here?

    • one year ago
  26. IsTim
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    Yes. thank you very much.

    • one year ago
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