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have you done something called the D-operator?

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

in other words you may use implicit differentiation ...

so then it's chain rule

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

yes using natural logs

ok then use the same method and differnetiate both sides

so it would be -1-2x

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

no wait

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

take the ln of both sides...

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

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

ok

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

now can you diff both sides implicitly ...

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

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

you used the product rule but u made a small error

let's work on this step by step

ok sure

now the editor is working

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

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

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

you mean + not =

yes

ok all you are left with is y x RHS

take the y to the other side

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

can you see the y?

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

the next question is what is y=?

just change the y to the main equation

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

no

delete that post

we don't have limits

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

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

yes, i got that

good let's us wrap it up

that is a different question now

the answer would be 1/e

it is easy to see

Please show me how it is easy to see.

sorry the answer would be e

ok to make life esear

you should use LH rule

let me prove that for you

yes, I need to understand this.

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

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

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

i actually just took a quiz on this

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

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

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

im prob wrong tho lol

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

can you see what will happen

(1/1+x)/x

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

yes plug in the 0

oh yes, 1

sorry

ln y =1

y =e

hehehehe

I see!

its a clever qs

then it will*

y = e^1

which means y =e

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

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

well good luck and all the best

Thank you!

you're welcome

where is my medal hehehe