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

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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 ...

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Other answers:

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...
use natural logs rules to obtain: ln(y)=-xln(1+x)
now can you diff both sides implicitly ...
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
\[ \ln(y)=-xln(1+x)\]
\[d( \ln(y)=-xln(1+x))\]
you mean + not =
ok all you are left with is y x RHS
take the y to the other side
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
delete that post
we don't have limits
can u see i changed the y to (1+x)^(1/x)
yes, i got that
good let's us wrap it up
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
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...
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
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
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
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
\[\ln(y) = \lim_{x \rightarrow 0} [\frac{\frac{1}{[ (1+x)}}{1}]\]
yes plug in the 0
oh yes, 1
ln y =1
y =e
I see!
its a clever qs
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 ...
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

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