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FreeTrader

  • 3 years ago

I am stuck with an easy problem - I must be missing something. I am trying to use the limit process to find the derivative of f(x) = 8 / sqrt of x. Can you help me see where I am messing up?

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  1. FreeTrader
    • 3 years ago
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    \[f(x) = 8/\sqrt{x}\] find the derivative using the limit process

  2. FreeTrader
    • 3 years ago
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    f'(x) = lim \[f^\prime(x) =\lim_{\Delta x \rightarrow 0} [(8/\sqrt{x+c}- 8/\sqrt{x})/c]\]

  3. FreeTrader
    • 3 years ago
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    I am banging my head against this problem and getting a bad answer every time. We can tell by looking at it that the answer is\[-(4/x^(3/2))\]

  4. FreeTrader
    • 3 years ago
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    Well, I think you see what I mean...

  5. FreeTrader
    • 3 years ago
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    But I haven't figured out the algebra to leave me with a 4 in the numerator. I keep getting stuck with things that don't work out in the end.

  6. FreeTrader
    • 3 years ago
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    Weeku, what do you have for me so far?

  7. Weeku
    • 3 years ago
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    Whoops, I made a mistake. Darn editor. . . Hold on.

  8. FreeTrader
    • 3 years ago
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    thanks

  9. FreeTrader
    • 3 years ago
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    Seems no matter what I try, I end up with delta x /[( sqrt x + delta x) + sqrt x]

  10. FreeTrader
    • 3 years ago
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    I am going in circles.

  11. FreeTrader
    • 3 years ago
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    are you still there?

  12. Weeku
    • 3 years ago
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    Yeah. I'm stuck as well. Hrm. . . I'm still trying, though. Sorry xD Interesting problem, though.

  13. Weeku
    • 3 years ago
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    Ah. Got it. Took a whole loose-leaf paper to work out. . . Pretty nasty.

  14. Weeku
    • 3 years ago
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    Did you get up to \[(8\sqrt{x} - 8\sqrt{x+h}) / (h \sqrt{x}\sqrt{x+h})\]? Or, do you know how I got there? At that point, multiple the numerator and denominator by the conjugate of the numerator.

  15. Weeku
    • 3 years ago
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    multiply*

  16. Weeku
    • 3 years ago
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    Then it becomes a lot of distributing and factoring, until you can finally divide out the h. Once that happens, substitute 0 for h. Phew. I dunno if I can type out everything I did. . . I probably did it the long way, but I saw no other, shorter way. Do you want me to type it out?

  17. FreeTrader
    • 3 years ago
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    Yes, I got that far. I see you have h = delta x and you canceled out one h already.

  18. FreeTrader
    • 3 years ago
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    I would be grateful if you typed it out.

  19. Weeku
    • 3 years ago
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    Before that, all I did was find the common denominator of 8/sqrt[x+h] and 8/sqrt[x], add them together, and instead of dividing by h, multiply by 1/h to get \[8\sqrt{x} - 8\sqrt{x+h} / h \sqrt{x} \sqrt{x+h}\] Then I factored out the 8 to get the above ^ Oh. Okay.

  20. Weeku
    • 3 years ago
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    Whoops. Never mind about factoring out the 8.

  21. Weeku
    • 3 years ago
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    I tried doing that, with no luck. :D

  22. Ishaan94
    • 3 years ago
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    \[\frac{\frac{8}{\sqrt{x+h}} - \frac{8}{\sqrt{x}}}{h}\]Firs Only the Numerator.... \[8\frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x+h}\times \sqrt{x}}\] Rationalization of Numerator \[\frac{-8}{\sqrt{x+h}\times\sqrt{x}\times(\sqrt{x} + \sqrt{x+h})}\] \[\frac{-8}{2\times x^{\frac{3}{2}}}\]

  23. FreeTrader
    • 3 years ago
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    Ishaan94, I am looking over what you have done to see where I diverged. Thank you.

  24. Weeku
    • 3 years ago
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    I lost you after "Rationalization of Numerator," but nicely done. I did it the very, very long way, I see. . . Sorry about that, FreeTrader.

  25. Ishaan94
    • 3 years ago
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    \[\Huge{\frac{\frac{-8\cancel{h}}{\sqrt{x+h}\times\sqrt{x}\times(\sqrt{x} + \sqrt{x+h})}}{\cancel{h}}}\]

  26. Ishaan94
    • 3 years ago
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    \[8\frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x} \times \sqrt{x+h}} \times \frac{\sqrt{x} + \sqrt{x+h}}{\sqrt{x} + \sqrt{x+h}}\]

  27. FreeTrader
    • 3 years ago
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    I see how Ishaan94 got to the "first only the numerator..." expression. But I am lost on how he got rid of the terms in the numerator.

  28. FreeTrader
    • 3 years ago
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    Right, multiply by conjugate.

  29. chaise
    • 3 years ago
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    I really think you are going about this the wrong way. You should change the square root into a power (1/2) and then bring the denominator to the top by changing the power to negative. Then you can simply use the limit process of: f(x)= 8x^(-1/2) f(x+h)=8(x+h)^(1/2) And plug this into the derivative formula. You should get rid of the terms in the denominator.

  30. chaise
    • 3 years ago
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    Then factor out a h and cancel.

  31. Ishaan94
    • 3 years ago
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    \[8\frac{x - (x+h)}{\sqrt{x}\times \sqrt{x+h} \times (\sqrt{x} + \sqrt{x+h})}\]

  32. FreeTrader
    • 3 years ago
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    I think it's dawning on me that I missed distributing the -1 on the h, too.

  33. FreeTrader
    • 3 years ago
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    Stick with me, I am going to check that on my paper.

  34. FreeTrader
    • 3 years ago
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    Ishaan94, when I perform the operation 8 *[ x -(x+h) ] I am left with 8x -8(x+h) which gives 8x -8x -8h

  35. Ishaan94
    • 3 years ago
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    Right

  36. Weeku
    • 3 years ago
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    Which gives -8h. divide out the h's.

  37. FreeTrader
    • 3 years ago
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    I am onto this now... working the algebra. searching for the lightswitch for the bulb over my head...

  38. chaise
    • 3 years ago
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    \[f(x)=\frac{8}{\sqrt{x}}\] \[f(x)=\frac{8}{x ^{1/2}}\] \[f(x)=8x ^{\frac{-1}{2}}\] \[f(x+h)=8(x+h)^{-1/2}\] \[f \prime(x)\lim_{h \rightarrow 0}=I can't remember the formula\] Now just use you derivative formula, cancel where necisary, factor out a h, then cancel with the denominator.

  39. Weeku
    • 3 years ago
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    Lol@chaise xD

  40. Ishaan94
    • 3 years ago
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    Lol

  41. FreeTrader
    • 3 years ago
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    My problem is now in the denominator of the numerator. I don't see a h to cancel the one against the -8.

  42. Weeku
    • 3 years ago
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    You're forgetting that the entire equation is over h at this point. \[\Huge{\frac{\frac{-8\cancel{h}}{\sqrt{x+h}\times\sqrt{x}\times(\sqrt{x} + \sqrt{x+h})}}{\cancel{h}}}\]

  43. FreeTrader
    • 3 years ago
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    I have that now. Thanks. I am down to working with the remaining denom terms.

  44. chaise
    • 3 years ago
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    All terms on the right hand side of the minus sign in your derivative formula that have a x term should cancel with the terms. I think you're going about this the wrong way. Instead of having a quotient you should really bring the denominator to the top by making the power negative. Remember this rule? am^-x = a/m^x

  45. FreeTrader
    • 3 years ago
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    Yes. I am already so deep in the prob that I am going to finish it with frax.

  46. FreeTrader
    • 3 years ago
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    Here's what I have in the denominator, which I am tempted to misspell as demon-ator. \[\sqrt{x} * \sqrt{x} * \sqrt{x+h} + \sqrt{x} * \sqrt{x+h} * \sqrt{x+h} \]

  47. FreeTrader
    • 3 years ago
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    Since the h values go to zero in the limit, the following is the denominator in the limit: \[2\sqrt{x}\sqrt{x}\sqrt{x}\]

  48. chaise
    • 3 years ago
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    \[f(x)=\frac{8}{\sqrt{x}}\] \[f(x)=\frac{8}{x ^{1/2}}\] \[f(x)=8x ^{\frac{-1}{2}}\] \[f(x+h)=8(x+h)^{-1/2}\] \[\lim_{h \rightarrow 0}f'(x)=\frac{f(x+h)-f(x)}{h}\] lim{h-->0) of f'(x) = (8(x+h)^(-1/2) - 8x^(-1/2))/h You can do the rest, I'm sure.

  49. Weeku
    • 3 years ago
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    Yes, freetrader, (sqrt[x])^3 = x^(3/2)

  50. FreeTrader
    • 3 years ago
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    Okay, I am losing my sanity. Can you help me remember if \[\sqrt{x^3} = \sqrt{x}\sqrt{x}\sqrt{x}\] ???

  51. chaise
    • 3 years ago
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    Yes - and this should be in the denominator of the equation once it has been solved. Notice how the ^(-1/2) becomes 1.5 when you increase the power by one. And yes it does FreeTrader

  52. Weeku
    • 3 years ago
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    Yup. That is a true statement.

  53. chaise
    • 3 years ago
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    Stop writing it as a square root until you get to the final stage of the question. sqrt(x)^3 = x^(3/2)

  54. FreeTrader
    • 3 years ago
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    I have temp insanity.

  55. Weeku
    • 3 years ago
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    'Tis okay.:)

  56. FreeTrader
    • 3 years ago
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    So, my answer took me 7 hours to reach.

  57. Weeku
    • 3 years ago
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    It was totally worth it!

  58. FreeTrader
    • 3 years ago
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    THANK YOU FRIENDS!

  59. Weeku
    • 3 years ago
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    :)

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