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

- anonymous

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

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- anonymous

have you done something called the D-operator?

- anonymous

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

- anonymous

in other words you may use implicit differentiation ...

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- anonymous

so then it's chain rule

- anonymous

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

- anonymous

yes using natural logs

- anonymous

ok then use the same method and differnetiate both sides

- anonymous

so it would be -1-2x

- anonymous

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

- anonymous

no wait

- anonymous

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

- anonymous

take the ln of both sides...

- anonymous

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

- anonymous

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

- anonymous

ok

- anonymous

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

- anonymous

now can you diff both sides implicitly ...

- anonymous

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

- anonymous

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

- anonymous

you used the product rule but u made a small error

- anonymous

let's work on this step by step

- anonymous

ok sure

- anonymous

now the editor is working

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

you mean + not =

- anonymous

yes

- anonymous

ok all you are left with is y x RHS

- anonymous

take the y to the other side

- anonymous

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

- anonymous

can you see the y?

- anonymous

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

- anonymous

the next question is what is y=?

- anonymous

just change the y to the main equation

- anonymous

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

- anonymous

no

- anonymous

delete that post

- anonymous

we don't have limits

- anonymous

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

- anonymous

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

- anonymous

yes, i got that

- anonymous

good let's us wrap it up

- 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

- 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...

- 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

- anonymous

that is a different question now

- anonymous

the answer would be 1/e

- anonymous

it is easy to see

- anonymous

Please show me how it is easy to see.

- anonymous

sorry the answer would be e

- 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

- anonymous

ok to make life esear

- anonymous

you should use LH rule

- anonymous

let me prove that for you

- anonymous

yes, I need to understand this.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

i actually just took a quiz on this

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

im prob wrong tho lol

- anonymous

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

- anonymous

can you see what will happen

- anonymous

(1/1+x)/x

- anonymous

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

- anonymous

yes plug in the 0

- anonymous

oh yes, 1

- anonymous

sorry

- anonymous

ln y =1

- anonymous

y =e

- anonymous

hehehehe

- anonymous

I see!

- anonymous

its a clever qs

- 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 ...

- anonymous

then it will*

- anonymous

y = e^1

- anonymous

which means y =e

- anonymous

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

- anonymous

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

- anonymous

well good luck and all the best

- anonymous

Thank you!

- anonymous

you're welcome

- anonymous

where is my medal hehehe

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