Question

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

Answer 1

have you done something called the D-operator?

Answer 2

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

Answer 3

in other words you may use implicit differentiation ...

Answer 4

so then it's chain rule

Answer 5

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

Answer 6

yes using natural logs

Answer 7

ok then use the same method and differnetiate both sides

Answer 8

so it would be -1-2x

Answer 9

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

Answer 10

no wait

Answer 11

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

Answer 12

take the ln of both sides...

Answer 13

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

Answer 14

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

Answer 15

ok

Answer 16

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

Answer 17

now can you diff both sides implicitly ...

Answer 18

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

Answer 19

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

Answer 20

you used the product rule but u made a small error

Answer 21

let's work on this step by step

Answer 22

ok sure

Answer 23

now the editor is working

Answer 24

\[ \ln(y)=-xln(1+x)\]

Answer 25

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

Answer 26

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

Answer 27

you mean + not =

Answer 28

yes

Answer 29

ok all you are left with is y x RHS

Answer 30

take the y to the other side

Answer 31

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

Answer 32

can you see the y?

Answer 33

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

Answer 34

the next question is what is y=?

Answer 35

just change the y to the main equation

Answer 36

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

Answer 37

no

Answer 38

delete that post

Answer 39

we don't have limits

Answer 40

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

Answer 41

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

Answer 42

yes, i got that

Answer 43

good let's us wrap it up

Answer 44

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

Answer 45

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...

Answer 46

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

Answer 47

that is a different question now

Answer 48

the answer would be 1/e

Answer 49

it is easy to see

Answer 50

Please show me how it is easy to see.

Answer 51

sorry the answer would be e

Answer 52

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

Answer 53

ok to make life esear

Answer 54

you should use LH rule

Answer 55

let me prove that for you

Answer 56

yes, I need to understand this.

Answer 57

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

Answer 58

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

Answer 59

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

Answer 60

i actually just took a quiz on this

Answer 61

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

Answer 62

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

Answer 63

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

Answer 64

im prob wrong tho lol

Answer 65

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

Answer 66

can you see what will happen

Answer 67

(1/1+x)/x

Answer 68

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

Answer 69

yes plug in the 0

Answer 70

oh yes, 1

Answer 71

sorry

Answer 72

ln y =1

Answer 73

y =e

Answer 74

hehehehe

Answer 75

I see!

Answer 76

its a clever qs

Answer 77

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 ...

Answer 78

then it will*

Answer 79

y = e^1

Answer 80

which means y =e

Answer 81

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

Answer 82

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

Answer 83

well good luck and all the best

Answer 84

Thank you!

Answer 85

you're welcome

Answer 86

where is my medal hehehe