anonymous
  • anonymous
Derivative of (1+x)^(1/x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
have you done something called the D-operator?
anonymous
  • anonymous
first of all rewrite that as (1+x)^(-x), does that clue or hint help?
anonymous
  • anonymous
in other words you may use implicit differentiation ...

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anonymous
  • anonymous
so then it's chain rule
anonymous
  • anonymous
ok just for you to never forget do you know how to find dy/dx of x^x?
anonymous
  • anonymous
yes using natural logs
anonymous
  • anonymous
ok then use the same method and differnetiate both sides
anonymous
  • anonymous
so it would be -1-2x
anonymous
  • anonymous
impossible to give that answer life is not linear...
anonymous
  • anonymous
no wait
anonymous
  • anonymous
y = (1+x)^(-x)
anonymous
  • anonymous
take the ln of both sides...
anonymous
  • anonymous
ln(y) = ln[(1+x)^(-x)]
anonymous
  • anonymous
my appology that the equation editor is not functioning for me tonight for a funny reason...
anonymous
  • anonymous
ok
anonymous
  • anonymous
use natural logs rules to obtain: ln(y)=-xln(1+x)
anonymous
  • anonymous
now can you diff both sides implicitly ...
anonymous
  • anonymous
f'(x)=-ln(1+x)-[x/(1+x)]*(1+x)^(-x)
anonymous
  • anonymous
no look at what i wrote again ln(y)=-xln(1+x)
anonymous
  • anonymous
you used the product rule but u made a small error
anonymous
  • anonymous
let's work on this step by step
anonymous
  • anonymous
ok sure
anonymous
  • anonymous
now the editor is working
anonymous
  • anonymous
\[ \ln(y)=-xln(1+x)\]
anonymous
  • anonymous
\[d( \ln(y)=-xln(1+x))\]
anonymous
  • anonymous
(1/y)y'=-1*ln(x+1)=(-x)(1/(1+x))
anonymous
  • anonymous
you mean + not =
anonymous
  • anonymous
yes
anonymous
  • anonymous
ok all you are left with is y x RHS
anonymous
  • anonymous
take the y to the other side
anonymous
  • anonymous
y'=y[-1*ln(x+1)=(-x)(1/(1+x))]
anonymous
  • anonymous
can you see the y?
anonymous
  • anonymous
yes... everything** is multiplied by y, not just tacked onto the end
anonymous
  • anonymous
the next question is what is y=?
anonymous
  • anonymous
just change the y to the main equation
anonymous
  • anonymous
now I have to apply that to x -> 0, L'Hopital's Rule
anonymous
  • anonymous
no
anonymous
  • anonymous
delete that post
anonymous
  • anonymous
we don't have limits
anonymous
  • anonymous
\[y'=[(1+x)^(1/x)][-1*\ln(x+1)=(-x)(1/(1+x))]\]
anonymous
  • anonymous
can u see i changed the y to (1+x)^(1/x)
anonymous
  • anonymous
yes, i got that
anonymous
  • anonymous
good let's us wrap it up
anonymous
  • 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
  • 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
  • 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
  • anonymous
that is a different question now
anonymous
  • anonymous
the answer would be 1/e
anonymous
  • anonymous
it is easy to see
anonymous
  • anonymous
Please show me how it is easy to see.
anonymous
  • anonymous
sorry the answer would be e
anonymous
  • 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
  • anonymous
ok to make life esear
anonymous
  • anonymous
you should use LH rule
anonymous
  • anonymous
let me prove that for you
anonymous
  • anonymous
yes, I need to understand this.
anonymous
  • anonymous
\[y = (1+x)^{(1/x)}\]
anonymous
  • anonymous
\[\ln(y) =\ln[ (1+x)^{(1/x)}]\]
anonymous
  • anonymous
\[\ln(y) =\frac{\ln[ (1+x)}{x}\]
anonymous
  • anonymous
i actually just took a quiz on this
anonymous
  • anonymous
are you at hopkins by chance? haha... I want to understand this
anonymous
  • anonymous
\[\ln(y) = \lim_{x \rightarrow 0} [\frac{\ln[ (1+x)}{x}]\]
anonymous
  • anonymous
i say (1+x)^1/x-1/x
anonymous
  • anonymous
im prob wrong tho lol
anonymous
  • anonymous
now differentiate the top and the butoom using LH rule...
anonymous
  • anonymous
can you see what will happen
anonymous
  • anonymous
(1/1+x)/x
anonymous
  • anonymous
\[\ln(y) = \lim_{x \rightarrow 0} [\frac{\frac{1}{[ (1+x)}}{1}]\]
anonymous
  • anonymous
yes plug in the 0
anonymous
  • anonymous
oh yes, 1
anonymous
  • anonymous
sorry
anonymous
  • anonymous
ln y =1
anonymous
  • anonymous
y =e
anonymous
  • anonymous
hehehehe
anonymous
  • anonymous
I see!
anonymous
  • anonymous
its a clever qs
anonymous
  • 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
  • anonymous
then it will*
anonymous
  • anonymous
y = e^1
anonymous
  • anonymous
which means y =e
anonymous
  • anonymous
did you get that...or am i disconnected...
anonymous
  • anonymous
no, i see that... much more advanced that what I'm used to seeing, but it makes sense.
anonymous
  • anonymous
well good luck and all the best
anonymous
  • anonymous
Thank you!
anonymous
  • anonymous
you're welcome
anonymous
  • anonymous
where is my medal hehehe

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