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IsTim

  • 2 years ago

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

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  1. IsTim
    • 2 years ago
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    This is as far as I got:|dw:1340139301477:dw|

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

  3. IsTim
    • 2 years ago
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    Ok.

  4. mashe
    • 2 years ago
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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

  5. IsTim
    • 2 years ago
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    How do I know a=0?

  6. mashe
    • 2 years ago
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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

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

  8. mashe
    • 2 years ago
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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

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

  10. IsTim
    • 2 years ago
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    Hello?

  11. myininaya
    • 2 years ago
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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?

  12. myininaya
    • 2 years ago
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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.

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

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

  15. myininaya
    • 2 years ago
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    So that is what you did?

  16. IsTim
    • 2 years ago
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    Yes, that was my previous step before this.

  17. myininaya
    • 2 years ago
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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?

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

  19. myininaya
    • 2 years ago
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    what else cancels?

  20. myininaya
    • 2 years ago
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    on top!

  21. IsTim
    • 2 years ago
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    I forgot to cancel 1 and -1.

  22. myininaya
    • 2 years ago
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    yep! :)

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

  24. IsTim
    • 2 years ago
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    yes.

  25. myininaya
    • 2 years ago
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    Do you know what to do from here?

  26. IsTim
    • 2 years ago
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    Yes. thank you very much.

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