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mathsmindBest ResponseYou've already chosen the best response.0
have you done something called the Doperator?
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
first of all rewrite that as (1+x)^(x), does that clue or hint help?
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
in other words you may use implicit differentiation ...
 one year ago

MFinn9Best ResponseYou've already chosen the best response.0
so then it's chain rule
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
ok just for you to never forget do you know how to find dy/dx of x^x?
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
ok then use the same method and differnetiate both sides
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
impossible to give that answer life is not linear...
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
take the ln of both sides...
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
ln(y) = ln[(1+x)^(x)]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
my appology that the equation editor is not functioning for me tonight for a funny reason...
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
use natural logs rules to obtain: ln(y)=xln(1+x)
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
now can you diff both sides implicitly ...
 one year ago

MFinn9Best ResponseYou've already chosen the best response.0
f'(x)=ln(1+x)[x/(1+x)]*(1+x)^(x)
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
no look at what i wrote again ln(y)=xln(1+x)
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
you used the product rule but u made a small error
 one year ago

MFinn9Best ResponseYou've already chosen the best response.0
let's work on this step by step
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
now the editor is working
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
\[ \ln(y)=xln(1+x)\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
\[d( \ln(y)=xln(1+x))\]
 one year ago

MFinn9Best ResponseYou've already chosen the best response.0
(1/y)y'=1*ln(x+1)=(x)(1/(1+x))
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
ok all you are left with is y x RHS
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
take the y to the other side
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
y'=y[1*ln(x+1)=(x)(1/(1+x))]
 one year ago

MFinn9Best ResponseYou've already chosen the best response.0
yes... everything** is multiplied by y, not just tacked onto the end
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
the next question is what is y=?
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
just change the y to the main equation
 one year ago

MFinn9Best ResponseYou've already chosen the best response.0
now I have to apply that to x > 0, L'Hopital's Rule
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
we don't have limits
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
\[y'=[(1+x)^(1/x)][1*\ln(x+1)=(x)(1/(1+x))]\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
can u see i changed the y to (1+x)^(1/x)
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
good let's us wrap it up
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
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 year ago

mathsmindBest ResponseYou've already chosen the best response.0
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...
 one year ago

MFinn9Best ResponseYou've already chosen the best response.0
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
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
that is a different question now
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
the answer would be 1/e
 one year ago

MFinn9Best ResponseYou've already chosen the best response.0
Please show me how it is easy to see.
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
sorry the answer would be e
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
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
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
ok to make life esear
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
you should use LH rule
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
let me prove that for you
 one year ago

MFinn9Best ResponseYou've already chosen the best response.0
yes, I need to understand this.
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
\[y = (1+x)^{(1/x)}\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
\[\ln(y) =\ln[ (1+x)^{(1/x)}]\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
\[\ln(y) =\frac{\ln[ (1+x)}{x}\]
 one year ago

EphiloBest ResponseYou've already chosen the best response.0
i actually just took a quiz on this
 one year ago

MFinn9Best ResponseYou've already chosen the best response.0
are you at hopkins by chance? haha... I want to understand this
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
\[\ln(y) = \lim_{x \rightarrow 0} [\frac{\ln[ (1+x)}{x}]\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
now differentiate the top and the butoom using LH rule...
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
can you see what will happen
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
\[\ln(y) = \lim_{x \rightarrow 0} [\frac{\frac{1}{[ (1+x)}}{1}]\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
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 ...
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
did you get that...or am i disconnected...
 one year ago

MFinn9Best ResponseYou've already chosen the best response.0
no, i see that... much more advanced that what I'm used to seeing, but it makes sense.
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
well good luck and all the best
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
where is my medal hehehe
 one year ago
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