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MFinn9

  • 2 years ago

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

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  1. mathsmind
    • 2 years ago
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    have you done something called the D-operator?

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

  3. mathsmind
    • 2 years ago
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    in other words you may use implicit differentiation ...

  4. MFinn9
    • 2 years ago
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    so then it's chain rule

  5. mathsmind
    • 2 years 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
    • 2 years ago
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    yes using natural logs

  7. mathsmind
    • 2 years ago
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    ok then use the same method and differnetiate both sides

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

  9. mathsmind
    • 2 years ago
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    impossible to give that answer life is not linear...

  10. MFinn9
    • 2 years ago
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    no wait

  11. mathsmind
    • 2 years ago
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    y = (1+x)^(-x)

  12. mathsmind
    • 2 years ago
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    take the ln of both sides...

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

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

  15. MFinn9
    • 2 years ago
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    ok

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

  17. mathsmind
    • 2 years ago
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    now can you diff both sides implicitly ...

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

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

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

  21. MFinn9
    • 2 years ago
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    let's work on this step by step

  22. mathsmind
    • 2 years ago
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    ok sure

  23. mathsmind
    • 2 years ago
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    now the editor is working

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

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

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

  27. mathsmind
    • 2 years ago
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    you mean + not =

  28. MFinn9
    • 2 years ago
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    yes

  29. mathsmind
    • 2 years ago
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    ok all you are left with is y x RHS

  30. mathsmind
    • 2 years ago
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    take the y to the other side

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

  32. mathsmind
    • 2 years ago
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    can you see the y?

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

  34. mathsmind
    • 2 years ago
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    the next question is what is y=?

  35. mathsmind
    • 2 years ago
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    just change the y to the main equation

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

  37. mathsmind
    • 2 years ago
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    no

  38. mathsmind
    • 2 years ago
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    delete that post

  39. mathsmind
    • 2 years ago
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    we don't have limits

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

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

  42. MFinn9
    • 2 years ago
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    yes, i got that

  43. mathsmind
    • 2 years ago
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    good let's us wrap it up

  44. mathsmind
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    that is a different question now

  48. mathsmind
    • 2 years ago
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    the answer would be 1/e

  49. mathsmind
    • 2 years ago
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    it is easy to see

  50. MFinn9
    • 2 years ago
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    Please show me how it is easy to see.

  51. mathsmind
    • 2 years ago
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    sorry the answer would be e

  52. mathsmind
    • 2 years 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
    • 2 years ago
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    ok to make life esear

  54. mathsmind
    • 2 years ago
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    you should use LH rule

  55. mathsmind
    • 2 years ago
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    let me prove that for you

  56. MFinn9
    • 2 years ago
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    yes, I need to understand this.

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

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

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

  60. Ephilo
    • 2 years ago
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    i actually just took a quiz on this

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

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

  63. Ephilo
    • 2 years ago
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    i say (1+x)^1/x-1/x

  64. Ephilo
    • 2 years ago
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    im prob wrong tho lol

  65. mathsmind
    • 2 years ago
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    now differentiate the top and the butoom using LH rule...

  66. mathsmind
    • 2 years ago
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    can you see what will happen

  67. MFinn9
    • 2 years ago
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    (1/1+x)/x

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

  69. mathsmind
    • 2 years ago
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    yes plug in the 0

  70. MFinn9
    • 2 years ago
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    oh yes, 1

  71. MFinn9
    • 2 years ago
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    sorry

  72. mathsmind
    • 2 years ago
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    ln y =1

  73. mathsmind
    • 2 years ago
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    y =e

  74. mathsmind
    • 2 years ago
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    hehehehe

  75. MFinn9
    • 2 years ago
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    I see!

  76. mathsmind
    • 2 years ago
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    its a clever qs

  77. mathsmind
    • 2 years 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
    • 2 years ago
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    then it will*

  79. mathsmind
    • 2 years ago
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    y = e^1

  80. mathsmind
    • 2 years ago
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    which means y =e

  81. mathsmind
    • 2 years ago
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    did you get that...or am i disconnected...

  82. MFinn9
    • 2 years 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
    • 2 years ago
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    well good luck and all the best

  84. MFinn9
    • 2 years ago
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    Thank you!

  85. mathsmind
    • 2 years ago
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    you're welcome

  86. mathsmind
    • 2 years ago
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    where is my medal hehehe

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