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IsTim

Find the equation of the tangent to the curve at the given x value.

  • one year ago
  • one year ago

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  1. IsTim
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    |dw:1335261368315:dw|

    • one year ago
  2. IsTim
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    I got up to this:|dw:1335261401688:dw|

    • one year ago
  3. IsTim
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    |dw:1335261446849:dw|

    • one year ago
  4. chaise
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    Calculate the derivative of the equation, make the equation to equal, solve it and plug in your x value. Are you familiar with the quotient rule in derivatives?

    • one year ago
  5. IsTim
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    Deriatives is f'(x) right? we're not doing that yet.

    • one year ago
  6. IsTim
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    Methods we're using is substitution, factoring and the conjugate method.

    • one year ago
  7. IsTim
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    I'm using the Calculus Phobe videos from Calculus Help (my teacher showed me the website and told me to use that).

    • one year ago
  8. chaise
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    Yes, that's correct. So you're differentiating using the definition? If you take the derivative of the equation y=3/(x^2-1) you can find the slope. if: f(x)=3/(x^2-1) then f(x+h)=3/((x+h)^2-1) dy/dx = (f(x+h)-f(x))/h Can you do the rest?

    • one year ago
  9. chaise
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    dy/dx =lim (h-->0) (f(x+h)-f(x))/h rather.

    • one year ago
  10. IsTim
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    Yes. That's what I did.

    • one year ago
  11. IsTim
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    But I can't go any further. |dw:1335262087563:dw| That's the furthest I got up to.

    • one year ago
  12. chaise
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    Expand your brackets and factor out a h.

    • one year ago
  13. IsTim
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    The bottom one to? I was told not too.

    • one year ago
  14. chaise
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    Not sure - I'm not familiar with the long method, I just use the shortcut :)

    • one year ago
  15. Callisto
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    Can you take derivatives? Must you use first principle to find it?

    • one year ago
  16. IsTim
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    Derivatives is our next unit.

    • one year ago
  17. IsTim
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    I can't isolate or factor out anything. I must had made a mistake beforehand I guess.

    • one year ago
  18. IsTim
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    Anyone?

    • one year ago
  19. Callisto
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    I hope you don't mind me posting my solution here

    • one year ago
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  20. IsTim
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    I guess not, but it's notm much like my work.

    • one year ago
  21. Callisto
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    Oh shoot... I got some mistakes there! Wait

    • one year ago
  22. Callisto
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    Once again :(

    • one year ago
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  23. Callisto
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    Oh my dear!!! The final answer should be -4/3 ... not -2 as I've written :S

    • one year ago
  24. Callisto
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    Is your question \[\frac{3}{x^2-1}\] or \[\frac{3}{x^2+1}\]?

    • one year ago
  25. IsTim
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    First one.

    • one year ago
  26. IsTim
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    Maybe I should upload my work, and maybe you check for problems?

    • one year ago
  27. Callisto
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    Sure!

    • one year ago
  28. IsTim
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    IT's c.

    • one year ago
  29. IsTim
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    Starting from c) at Scan001.

    • one year ago
  30. Callisto
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    f(a) is not 1

    • one year ago
  31. Callisto
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    f(a) is 3/ (a^2 -1)

    • one year ago
  32. IsTim
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    but since x=2, I subbed in a=2 into that equation...

    • one year ago
  33. Callisto
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    for the denominator, (a+h)-a =h , not a+h -2 You need to find the limit before substituting the numbers.

    • one year ago
  34. Callisto
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    After you've found the limit, you'll see the unknown x. By then, sub x=2 and you'll get the slope. But NOT when you're taking the limit

    • one year ago
  35. IsTim
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    I thought the unknown to be m, and that h was still the variable withint the equation?

    • one year ago
  36. Callisto
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    h is something you need to deal with. Should I upload an example to should how you do first principle for you ?

    • one year ago
  37. IsTim
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    I've got an example here with me from my teacher.

    • one year ago
  38. Callisto
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    If you don't need it, it's okay :) Should I keep checking for you or you want to do corrections first?

    • one year ago
  39. IsTim
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    What should I correct?

    • one year ago
  40. Callisto
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    Correct starting from the third line For numerator: a -> 3/ (a^2 -1) For denominator: a+h-2 -> h (in your previous step, a+h -a, simplify it , you'll get h)

    • one year ago
  41. IsTim
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    Sorry if I sound ignorant, but that's how I saw it in the example...the h is alone.

    • one year ago
  42. Callisto
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    Do you mind uploading your example also?

    • one year ago
  43. IsTim
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    • one year ago
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  44. Callisto
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    First, the situation is different . You're given 2 values there, but only 1 here. Second, to be frank, I don't understand what you (or your teacher) is doing. It's not the normal way as I've learnt.. Third, I still insist that you shouldn't sub the value immediately...

    • one year ago
  45. IsTim
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    Ok. I think I know what you are doing, and what my teacher is doing. You're going from a general formula, while my teacher is using the specific formula. The general formula doesn't sub in the x or a value until the end.

    • one year ago
  46. Callisto
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    It's good that you notice the difference :)

    • one year ago
  47. IsTim
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    Just looking through my notes I guess. I've got a test coming up, so I thought reviewing would be best.

    • one year ago
  48. Callisto
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    Yup :)

    • one year ago
  49. Mimi_x3
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    well, where do you need help with? This is pretty long..

    • one year ago
  50. IsTim
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    Maybe I should just leave this question. I tried this for over 2 hours by now.

    • one year ago
  51. Mimi_x3
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    First, what is the question? \[\frac{1}{x-2} \]?? or \[\frac{3}{x^{2}-1} \]??

    • one year ago
  52. IsTim
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    Second one, whereas x=2.

    • one year ago
  53. IsTim
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    Ok. Thanks for the attempt guys. I'll try other problems, and maybe this might make more sense.

    • one year ago
  54. Mimi_x3
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    wait..i will try and do it first.. can't you take the derivative which is easier?

    • one year ago
  55. IsTim
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    Others suggested that, but that's for the next unit. I understand a bit on how that works, but I'm not allowed using it here.

    • one year ago
  56. Mimi_x3
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    Hmm..I might try it but it might take a while since i havent done these in a while...

    • one year ago
  57. IsTim
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    Ok. Well, I guess you can do so, if yo uhave free time. But most likely you too have something important coming up (immediately).

    • one year ago
  58. Mimi_x3
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    No, I don't have anything important..i might be able to do it but not sure..lol

    • one year ago
  59. Mimi_x3
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    I have to review these anyway

    • one year ago
  60. apoorvk
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    ^^That was mammoth. phew. Istim do you know how to differentiate using chain rule?

    • one year ago
  61. Mimi_x3
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    lol, i failed...do you remember how to do these apoor? my brain is failling me :(

    • one year ago
  62. IsTim
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    No, I don't (Sorry, cleaning up house).

    • one year ago
  63. apoorvk
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    Hmm. so you want to do this by the first principle of finding derivatives okay.

    • one year ago
  64. IsTim
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    I am just oh so selective in the method used. That be the end of me.

    • one year ago
  65. Mimi_x3
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    \[\large\frac{f(x+h)-f(x)}{h} =>\frac{\frac{3}{(x+h)^{2}-1}-\frac{3}{x^{2}-1}}{h} \]

    • one year ago
  66. Mimi_x3
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    right so far?

    • one year ago
  67. IsTim
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    I have no idea.

    • one year ago
  68. chaise
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    should be the limit as h approaches zero, but, apart from that, correct. ^_^

    • one year ago
  69. Mimi_x3
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    \[\large =>\frac{\frac{3}{h^{2}+2hx+x^{2}-1}-\frac{3}{x^{2}-1}}{h} \] \[\large >\frac{\frac{3(x^{2}-1)-3(h^{2}+2hx+x^{2}-1)}{(h^{2}+2hx+x^{2}-1)(x^{2}-1)}}{h} \]

    • one year ago
  70. Mimi_x3
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    I think that im doing it wrong..i dont think that its this complicated..

    • one year ago
  71. chaise
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    Perhaps you should just use quotient rule.

    • one year ago
  72. Mimi_x3
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    but it says derivatiive thing is not allowed..

    • one year ago
  73. IsTim
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    Before using deratives. Referencing Calculus Phobe from Calculus Help. I think I used Conjugate Method.

    • one year ago
  74. Mimi_x3
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    whaat..im lost..i looked at your example but it got confusing..then i looked at my book its more confusing..using the chain rule is easier.

    • one year ago
  75. Mimi_x3
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    \[\frac{3}{x^{2}-1} => 3(x^2-1)^{-1}\] \[-3(x^2-1)^{-1}*2x \] \[-6x(x^2-1)^{-1}\] easier i suppose..

    • one year ago
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