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anonymous

  • 5 years ago

using the definition of a derivative find f(x) =-2/sqrt{x}

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  1. anonymous
    • 5 years ago
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    \[f(x)= -2/ \sqrt{x}\]

  2. anonymous
    • 5 years ago
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    this will be a pain to write out but not hard to compute. you must use the definition yes?

  3. anonymous
    • 5 years ago
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    i know that part i am having problems with rationalizing the denominator

  4. anonymous
    • 5 years ago
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    you need to compute \[lim_{h->0}\frac{\frac{-2}{\sqrt{x+h}}+\frac{2}{\sqrt{x}}}{h}\]

  5. watchmath
    • 5 years ago
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    Good opportunity to practice your LaTeX satellite! :)

  6. anonymous
    • 5 years ago
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    go ahead watchmath.

  7. anonymous
    • 5 years ago
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    i am learning. eventually i will be fluent in it

  8. anonymous
    • 5 years ago
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    i know that part the step after that

  9. anonymous
    • 5 years ago
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    ok lets give me a break and at least factor out the -2, since this has nothing to do with the limit ok?

  10. anonymous
    • 5 years ago
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    just factor out by 2 thats all you had t say lol thanks so much

  11. anonymous
    • 5 years ago
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    i knew that i was just doubting myself

  12. anonymous
    • 5 years ago
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    i knew that i was just doubting myself

  13. anonymous
    • 5 years ago
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    so we will just write \[lim_{h->0}\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}\]

  14. anonymous
    • 5 years ago
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    all divided by h of course. we worry about that last.

  15. anonymous
    • 5 years ago
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    so now we rationalize

  16. anonymous
    • 5 years ago
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    \[\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}=\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x}\sqrt{x+h}}\]

  17. anonymous
    • 5 years ago
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    yes multiply numerator and denominator by the conjugate of th enumerator.

  18. anonymous
    • 5 years ago
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    which of course is \[\sqrt{x}+\sqrt{x+h}\]

  19. anonymous
    • 5 years ago
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    when you do that the numerator will just be \[x-x+h=h\]

  20. anonymous
    • 5 years ago
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    and the denominator will be the product \[\sqrt{x} \sqrt{x+h}) (\sqrt{x}+\sqrt{x+h})\]

  21. anonymous
    • 5 years ago
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    sorry am not repsonding i was working it out thanks i got it

  22. anonymous
    • 5 years ago
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    thats ok i was typing. did you get the numerator correctly? it is just h

  23. anonymous
    • 5 years ago
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    yes the x's cancel

  24. anonymous
    • 5 years ago
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    the denominator is that ugly thing i wrote last. so you have in total \[\frac{h}{h(\sqrt{x}\sqrt{x+h})(\sqrt{x}+\sqrt{x+h}}\]

  25. anonymous
    • 5 years ago
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    yes they 'cancel' meaning they add to zero. and of course dividing by h means the h goes in the denominator

  26. anonymous
    • 5 years ago
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    the h cancels

  27. anonymous
    • 5 years ago
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    so that last ugly thing i wrote is what you get when you rationalize the numerator. yes now the h cancels.

  28. anonymous
    • 5 years ago
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    leaving \[\frac{1}{(\sqrt{x}\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}\]

  29. anonymous
    • 5 years ago
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    now you can replace h by 0 since you will not be dividing by 0.

  30. anonymous
    • 5 years ago
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    to get... \[\frac{1}{(\sqrt{x}\sqrt{x})(\sqrt{x}+\sqrt{x})}\]

  31. anonymous
    • 5 years ago
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    yes

  32. anonymous
    • 5 years ago
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    \[=\frac{1}{2x\sqrt{x}}\]

  33. anonymous
    • 5 years ago
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    oh damn i made a mistake early on the numerator was \[x-(x+h)=-h\] not \[x-x+h\]

  34. anonymous
    • 5 years ago
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    a very bush league mistake but easily rectified. just replace the 1 in the numerator by -1

  35. anonymous
    • 5 years ago
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    sorry!

  36. anonymous
    • 5 years ago
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    so the correct answer is \[\frac{-1}{2x\sqrt{x}}\]

  37. anonymous
    • 5 years ago
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    then don't forget to multiply by the -2 at the end because we factored it out.

  38. anonymous
    • 5 years ago
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    so the "final answer" as we say is \[\frac{1}{x\sqrt{x}}\] sorry it took a while

  39. anonymous
    • 5 years ago
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    all steps clear?

  40. anonymous
    • 5 years ago
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    sorry my computer is freezing idk if its thes site or what

  41. anonymous
    • 5 years ago
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    sometime the site is funky. earlier today it certainly was.

  42. anonymous
    • 5 years ago
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    if you have a question about any step post and i will respond

  43. anonymous
    • 5 years ago
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    i understand thank you sooo much!

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