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MFinn9 Group Title

Derivative of (1+x)^(1/x)

  • one year ago
  • one year ago

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  1. mathsmind Group Title
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    have you done something called the D-operator?

    • one year ago
  2. mathsmind Group Title
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    first of all rewrite that as (1+x)^(-x), does that clue or hint help?

    • one year ago
  3. mathsmind Group Title
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    in other words you may use implicit differentiation ...

    • one year ago
  4. MFinn9 Group Title
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    so then it's chain rule

    • one year ago
  5. mathsmind Group Title
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    ok just for you to never forget do you know how to find dy/dx of x^x?

    • one year ago
  6. MFinn9 Group Title
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    yes using natural logs

    • one year ago
  7. mathsmind Group Title
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    ok then use the same method and differnetiate both sides

    • one year ago
  8. MFinn9 Group Title
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    so it would be -1-2x

    • one year ago
  9. mathsmind Group Title
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    impossible to give that answer life is not linear...

    • one year ago
  10. MFinn9 Group Title
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    no wait

    • one year ago
  11. mathsmind Group Title
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    y = (1+x)^(-x)

    • one year ago
  12. mathsmind Group Title
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    take the ln of both sides...

    • one year ago
  13. mathsmind Group Title
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    ln(y) = ln[(1+x)^(-x)]

    • one year ago
  14. mathsmind Group Title
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    my appology that the equation editor is not functioning for me tonight for a funny reason...

    • one year ago
  15. MFinn9 Group Title
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    ok

    • one year ago
  16. mathsmind Group Title
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    use natural logs rules to obtain: ln(y)=-xln(1+x)

    • one year ago
  17. mathsmind Group Title
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    now can you diff both sides implicitly ...

    • one year ago
  18. MFinn9 Group Title
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    f'(x)=-ln(1+x)-[x/(1+x)]*(1+x)^(-x)

    • one year ago
  19. mathsmind Group Title
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    no look at what i wrote again ln(y)=-xln(1+x)

    • one year ago
  20. mathsmind Group Title
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    you used the product rule but u made a small error

    • one year ago
  21. MFinn9 Group Title
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    let's work on this step by step

    • one year ago
  22. mathsmind Group Title
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    ok sure

    • one year ago
  23. mathsmind Group Title
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    now the editor is working

    • one year ago
  24. mathsmind Group Title
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    \[ \ln(y)=-xln(1+x)\]

    • one year ago
  25. mathsmind Group Title
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    \[d( \ln(y)=-xln(1+x))\]

    • one year ago
  26. MFinn9 Group Title
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    (1/y)y'=-1*ln(x+1)=(-x)(1/(1+x))

    • one year ago
  27. mathsmind Group Title
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    you mean + not =

    • one year ago
  28. MFinn9 Group Title
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    yes

    • one year ago
  29. mathsmind Group Title
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    ok all you are left with is y x RHS

    • one year ago
  30. mathsmind Group Title
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    take the y to the other side

    • one year ago
  31. mathsmind Group Title
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    y'=y[-1*ln(x+1)=(-x)(1/(1+x))]

    • one year ago
  32. mathsmind Group Title
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    can you see the y?

    • one year ago
  33. MFinn9 Group Title
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    yes... everything** is multiplied by y, not just tacked onto the end

    • one year ago
  34. mathsmind Group Title
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    the next question is what is y=?

    • one year ago
  35. mathsmind Group Title
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    just change the y to the main equation

    • one year ago
  36. MFinn9 Group Title
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    now I have to apply that to x -> 0, L'Hopital's Rule

    • one year ago
  37. mathsmind Group Title
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    no

    • one year ago
  38. mathsmind Group Title
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    delete that post

    • one year ago
  39. mathsmind Group Title
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    we don't have limits

    • one year ago
  40. mathsmind Group Title
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    \[y'=[(1+x)^(1/x)][-1*\ln(x+1)=(-x)(1/(1+x))]\]

    • one year ago
  41. mathsmind Group Title
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    can u see i changed the y to (1+x)^(1/x)

    • one year ago
  42. MFinn9 Group Title
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    yes, i got that

    • one year ago
  43. mathsmind Group Title
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    good let's us wrap it up

    • one year ago
  44. mathsmind Group Title
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    in such questions 1) you take the natrual log of both side, 2) you apply the logs rules, 3) you implicitly differentiate both sides, 4) you plug in the oridinal equation or function back

    • one year ago
  45. mathsmind Group Title
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    one of the interesting things about this phenomina, that you'll have the same function muliplied by it derivative in the logarithmic world, a lot of real life applications can be modeled by this system...

    • one year ago
  46. MFinn9 Group Title
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    Ok, great... now the whole point of that was to answer the question: lim x->0 (1+0)^(1/x), and applying l'hopital's rule

    • one year ago
  47. mathsmind Group Title
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    that is a different question now

    • one year ago
  48. mathsmind Group Title
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    the answer would be 1/e

    • one year ago
  49. mathsmind Group Title
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    it is easy to see

    • one year ago
  50. MFinn9 Group Title
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    Please show me how it is easy to see.

    • one year ago
  51. mathsmind Group Title
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    sorry the answer would be e

    • one year ago
  52. mathsmind Group Title
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    now one way to visualise this is by graphical method and the definition of limits, the other way to find out is by using calculus normal methods

    • one year ago
  53. mathsmind Group Title
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    ok to make life esear

    • one year ago
  54. mathsmind Group Title
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    you should use LH rule

    • one year ago
  55. mathsmind Group Title
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    let me prove that for you

    • one year ago
  56. MFinn9 Group Title
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    yes, I need to understand this.

    • one year ago
  57. mathsmind Group Title
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    \[y = (1+x)^{(1/x)}\]

    • one year ago
  58. mathsmind Group Title
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    \[\ln(y) =\ln[ (1+x)^{(1/x)}]\]

    • one year ago
  59. mathsmind Group Title
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    \[\ln(y) =\frac{\ln[ (1+x)}{x}\]

    • one year ago
  60. Ephilo Group Title
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    i actually just took a quiz on this

    • one year ago
  61. MFinn9 Group Title
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    are you at hopkins by chance? haha... I want to understand this

    • one year ago
  62. mathsmind Group Title
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    \[\ln(y) = \lim_{x \rightarrow 0} [\frac{\ln[ (1+x)}{x}]\]

    • one year ago
  63. Ephilo Group Title
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    i say (1+x)^1/x-1/x

    • one year ago
  64. Ephilo Group Title
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    im prob wrong tho lol

    • one year ago
  65. mathsmind Group Title
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    now differentiate the top and the butoom using LH rule...

    • one year ago
  66. mathsmind Group Title
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    can you see what will happen

    • one year ago
  67. MFinn9 Group Title
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    (1/1+x)/x

    • one year ago
  68. mathsmind Group Title
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    \[\ln(y) = \lim_{x \rightarrow 0} [\frac{\frac{1}{[ (1+x)}}{1}]\]

    • one year ago
  69. mathsmind Group Title
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    yes plug in the 0

    • one year ago
  70. MFinn9 Group Title
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    oh yes, 1

    • one year ago
  71. MFinn9 Group Title
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    sorry

    • one year ago
  72. mathsmind Group Title
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    ln y =1

    • one year ago
  73. mathsmind Group Title
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    y =e

    • one year ago
  74. mathsmind Group Title
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    hehehehe

    • one year ago
  75. MFinn9 Group Title
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    I see!

    • one year ago
  76. mathsmind Group Title
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    its a clever qs

    • one year ago
  77. mathsmind Group Title
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    this is the first time is see such a problem, but just follow the laws of maths, the it will drive us mad and crazy ...

    • one year ago
  78. mathsmind Group Title
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    then it will*

    • one year ago
  79. mathsmind Group Title
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    y = e^1

    • one year ago
  80. mathsmind Group Title
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    which means y =e

    • one year ago
  81. mathsmind Group Title
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    did you get that...or am i disconnected...

    • one year ago
  82. MFinn9 Group Title
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    no, i see that... much more advanced that what I'm used to seeing, but it makes sense.

    • one year ago
  83. mathsmind Group Title
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    well good luck and all the best

    • one year ago
  84. MFinn9 Group Title
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    Thank you!

    • one year ago
  85. mathsmind Group Title
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    you're welcome

    • one year ago
  86. mathsmind Group Title
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    where is my medal hehehe

    • one year ago
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