## MFinn9 2 years ago Derivative of (1+x)^(1/x)

1. mathsmind

have you done something called the D-operator?

2. mathsmind

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

3. mathsmind

in other words you may use implicit differentiation ...

4. MFinn9

so then it's chain rule

5. mathsmind

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

6. MFinn9

yes using natural logs

7. mathsmind

ok then use the same method and differnetiate both sides

8. MFinn9

so it would be -1-2x

9. mathsmind

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

10. MFinn9

no wait

11. mathsmind

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

12. mathsmind

take the ln of both sides...

13. mathsmind

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

14. mathsmind

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

15. MFinn9

ok

16. mathsmind

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

17. mathsmind

now can you diff both sides implicitly ...

18. MFinn9

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

19. mathsmind

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

20. mathsmind

you used the product rule but u made a small error

21. MFinn9

let's work on this step by step

22. mathsmind

ok sure

23. mathsmind

now the editor is working

24. mathsmind

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

25. mathsmind

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

26. MFinn9

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

27. mathsmind

you mean + not =

28. MFinn9

yes

29. mathsmind

ok all you are left with is y x RHS

30. mathsmind

take the y to the other side

31. mathsmind

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

32. mathsmind

can you see the y?

33. MFinn9

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

34. mathsmind

the next question is what is y=?

35. mathsmind

just change the y to the main equation

36. MFinn9

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

37. mathsmind

no

38. mathsmind

delete that post

39. mathsmind

we don't have limits

40. mathsmind

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

41. mathsmind

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

42. MFinn9

yes, i got that

43. mathsmind

good let's us wrap it up

44. mathsmind

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 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

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

that is a different question now

48. mathsmind

the answer would be 1/e

49. mathsmind

it is easy to see

50. MFinn9

Please show me how it is easy to see.

51. mathsmind

sorry the answer would be e

52. mathsmind

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

ok to make life esear

54. mathsmind

you should use LH rule

55. mathsmind

let me prove that for you

56. MFinn9

yes, I need to understand this.

57. mathsmind

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

58. mathsmind

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

59. mathsmind

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

60. Ephilo

i actually just took a quiz on this

61. MFinn9

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

62. mathsmind

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

63. Ephilo

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

64. Ephilo

im prob wrong tho lol

65. mathsmind

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

66. mathsmind

can you see what will happen

67. MFinn9

(1/1+x)/x

68. mathsmind

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

69. mathsmind

yes plug in the 0

70. MFinn9

oh yes, 1

71. MFinn9

sorry

72. mathsmind

ln y =1

73. mathsmind

y =e

74. mathsmind

hehehehe

75. MFinn9

I see!

76. mathsmind

its a clever qs

77. mathsmind

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

then it will*

79. mathsmind

y = e^1

80. mathsmind

which means y =e

81. mathsmind

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

82. MFinn9

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

83. mathsmind

well good luck and all the best

84. MFinn9

Thank you!

85. mathsmind

you're welcome

86. mathsmind

where is my medal hehehe