## anonymous 3 years ago Derivative of (1+x)^(1/x)

1. anonymous

have you done something called the D-operator?

2. anonymous

first of all rewrite that as (1+x)^(-x), does that clue or hint help?

3. anonymous

in other words you may use implicit differentiation ...

4. anonymous

so then it's chain rule

5. anonymous

ok just for you to never forget do you know how to find dy/dx of x^x?

6. anonymous

yes using natural logs

7. anonymous

ok then use the same method and differnetiate both sides

8. anonymous

so it would be -1-2x

9. anonymous

impossible to give that answer life is not linear...

10. anonymous

no wait

11. anonymous

y = (1+x)^(-x)

12. anonymous

take the ln of both sides...

13. anonymous

ln(y) = ln[(1+x)^(-x)]

14. anonymous

my appology that the equation editor is not functioning for me tonight for a funny reason...

15. anonymous

ok

16. anonymous

use natural logs rules to obtain: ln(y)=-xln(1+x)

17. anonymous

now can you diff both sides implicitly ...

18. anonymous

f'(x)=-ln(1+x)-[x/(1+x)]*(1+x)^(-x)

19. anonymous

no look at what i wrote again ln(y)=-xln(1+x)

20. anonymous

you used the product rule but u made a small error

21. anonymous

let's work on this step by step

22. anonymous

ok sure

23. anonymous

now the editor is working

24. anonymous

$\ln(y)=-xln(1+x)$

25. anonymous

$d( \ln(y)=-xln(1+x))$

26. anonymous

(1/y)y'=-1*ln(x+1)=(-x)(1/(1+x))

27. anonymous

you mean + not =

28. anonymous

yes

29. anonymous

ok all you are left with is y x RHS

30. anonymous

take the y to the other side

31. anonymous

y'=y[-1*ln(x+1)=(-x)(1/(1+x))]

32. anonymous

can you see the y?

33. anonymous

yes... everything** is multiplied by y, not just tacked onto the end

34. anonymous

the next question is what is y=?

35. anonymous

just change the y to the main equation

36. anonymous

now I have to apply that to x -> 0, L'Hopital's Rule

37. anonymous

no

38. anonymous

delete that post

39. anonymous

we don't have limits

40. anonymous

$y'=[(1+x)^(1/x)][-1*\ln(x+1)=(-x)(1/(1+x))]$

41. anonymous

can u see i changed the y to (1+x)^(1/x)

42. anonymous

yes, i got that

43. anonymous

good let's us wrap it up

44. anonymous

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. anonymous

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. anonymous

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. anonymous

that is a different question now

48. anonymous

49. anonymous

it is easy to see

50. anonymous

Please show me how it is easy to see.

51. anonymous

sorry the answer would be e

52. anonymous

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. anonymous

ok to make life esear

54. anonymous

you should use LH rule

55. anonymous

let me prove that for you

56. anonymous

yes, I need to understand this.

57. anonymous

$y = (1+x)^{(1/x)}$

58. anonymous

$\ln(y) =\ln[ (1+x)^{(1/x)}]$

59. anonymous

$\ln(y) =\frac{\ln[ (1+x)}{x}$

60. anonymous

i actually just took a quiz on this

61. anonymous

are you at hopkins by chance? haha... I want to understand this

62. anonymous

$\ln(y) = \lim_{x \rightarrow 0} [\frac{\ln[ (1+x)}{x}]$

63. anonymous

i say (1+x)^1/x-1/x

64. anonymous

im prob wrong tho lol

65. anonymous

now differentiate the top and the butoom using LH rule...

66. anonymous

can you see what will happen

67. anonymous

(1/1+x)/x

68. anonymous

$\ln(y) = \lim_{x \rightarrow 0} [\frac{\frac{1}{[ (1+x)}}{1}]$

69. anonymous

yes plug in the 0

70. anonymous

oh yes, 1

71. anonymous

sorry

72. anonymous

ln y =1

73. anonymous

y =e

74. anonymous

hehehehe

75. anonymous

I see!

76. anonymous

its a clever qs

77. anonymous

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. anonymous

then it will*

79. anonymous

y = e^1

80. anonymous

which means y =e

81. anonymous

did you get that...or am i disconnected...

82. anonymous

no, i see that... much more advanced that what I'm used to seeing, but it makes sense.

83. anonymous

well good luck and all the best

84. anonymous

Thank you!

85. anonymous

you're welcome

86. anonymous

where is my medal hehehe