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MFinn9

  • one year ago

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

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  1. mathsmind
    • one year ago
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    have you done something called the D-operator?

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

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

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

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

  6. MFinn9
    • one year ago
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    yes using natural logs

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

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

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

  10. MFinn9
    • one year ago
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    no wait

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

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

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

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

  15. MFinn9
    • one year ago
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    ok

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

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

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

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

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

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

  22. mathsmind
    • one year ago
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    ok sure

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

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

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

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

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

  28. MFinn9
    • one year ago
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    yes

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

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

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

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

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

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

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

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

  37. mathsmind
    • one year ago
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    no

  38. mathsmind
    • one year ago
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    delete that post

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

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

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

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

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

  44. mathsmind
    • one year ago
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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

  45. mathsmind
    • one year ago
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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...

  46. MFinn9
    • one year ago
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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

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

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

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

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

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

  52. mathsmind
    • one year ago
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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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  70. MFinn9
    • one year ago
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    oh yes, 1

  71. MFinn9
    • one year ago
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    sorry

  72. mathsmind
    • one year ago
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    ln y =1

  73. mathsmind
    • one year ago
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    y =e

  74. mathsmind
    • one year ago
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    hehehehe

  75. MFinn9
    • one year ago
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    I see!

  76. mathsmind
    • one year ago
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    its a clever qs

  77. mathsmind
    • one year ago
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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 ...

  78. mathsmind
    • one year ago
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    then it will*

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

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

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

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

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

  84. MFinn9
    • one year ago
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    Thank you!

  85. mathsmind
    • one year ago
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    you're welcome

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

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