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anonymous
 3 years ago
Derivative of (1+x)^(1/x)
anonymous
 3 years ago
Derivative of (1+x)^(1/x)

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0have you done something called the Doperator?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0in other words you may use implicit differentiation ...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so then it's chain rule

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes using natural logs

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok then use the same method and differnetiate both sides

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0impossible to give that answer life is not linear...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0take the ln of both sides...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ln(y) = ln[(1+x)^(x)]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0my appology that the equation editor is not functioning for me tonight for a funny reason...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0use natural logs rules to obtain: ln(y)=xln(1+x)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now can you diff both sides implicitly ...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0f'(x)=ln(1+x)[x/(1+x)]*(1+x)^(x)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no look at what i wrote again ln(y)=xln(1+x)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you used the product rule but u made a small error

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0let's work on this step by step

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now the editor is working

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \ln(y)=xln(1+x)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[d( \ln(y)=xln(1+x))\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0(1/y)y'=1*ln(x+1)=(x)(1/(1+x))

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok all you are left with is y x RHS

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0take the y to the other side

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0y'=y[1*ln(x+1)=(x)(1/(1+x))]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes... everything** is multiplied by y, not just tacked onto the end

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the next question is what is y=?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0just change the y to the main equation

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now I have to apply that to x > 0, L'Hopital's Rule

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[y'=[(1+x)^(1/x)][1*\ln(x+1)=(x)(1/(1+x))]\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can u see i changed the y to (1+x)^(1/x)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0good let's us wrap it up

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0in 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
 3 years ago
Best ResponseYou've already chosen the best response.0one 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
 3 years ago
Best ResponseYou've already chosen the best response.0Ok, 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
 3 years ago
Best ResponseYou've already chosen the best response.0that is a different question now

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the answer would be 1/e

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Please show me how it is easy to see.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry the answer would be e

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now 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
 3 years ago
Best ResponseYou've already chosen the best response.0ok to make life esear

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you should use LH rule

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0let me prove that for you

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes, I need to understand this.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[y = (1+x)^{(1/x)}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\ln(y) =\ln[ (1+x)^{(1/x)}]\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\ln(y) =\frac{\ln[ (1+x)}{x}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i actually just took a quiz on this

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0are you at hopkins by chance? haha... I want to understand this

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\ln(y) = \lim_{x \rightarrow 0} [\frac{\ln[ (1+x)}{x}]\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0im prob wrong tho lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now differentiate the top and the butoom using LH rule...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you see what will happen

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\ln(y) = \lim_{x \rightarrow 0} [\frac{\frac{1}{[ (1+x)}}{1}]\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this 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
 3 years ago
Best ResponseYou've already chosen the best response.0did you get that...or am i disconnected...

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well good luck and all the best

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0where is my medal hehehe
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