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

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

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

MFinn9
 one year ago
Best ResponseYou've already chosen the best response.0so then it's chain rule

mathsmind
 one year 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?

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

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

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

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

mathsmind
 one year 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...

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

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

MFinn9
 one year ago
Best ResponseYou've already chosen the best response.0f'(x)=ln(1+x)[x/(1+x)]*(1+x)^(x)

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

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

MFinn9
 one year ago
Best ResponseYou've already chosen the best response.0let's work on this step by step

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

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

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

MFinn9
 one year ago
Best ResponseYou've already chosen the best response.0(1/y)y'=1*ln(x+1)=(x)(1/(1+x))

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

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

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

MFinn9
 one year ago
Best ResponseYou've already chosen the best response.0yes... everything** is multiplied by y, not just tacked onto the end

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

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

MFinn9
 one year ago
Best ResponseYou've already chosen the best response.0now I have to apply that to x > 0, L'Hopital's Rule

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

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

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

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

mathsmind
 one year 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

mathsmind
 one year 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...

MFinn9
 one year 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

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

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

MFinn9
 one year ago
Best ResponseYou've already chosen the best response.0Please show me how it is easy to see.

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

mathsmind
 one year 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

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

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

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

MFinn9
 one year ago
Best ResponseYou've already chosen the best response.0yes, I need to understand this.

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

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

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

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

MFinn9
 one year ago
Best ResponseYou've already chosen the best response.0are you at hopkins by chance? haha... I want to understand this

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

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

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

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

mathsmind
 one year 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 ...

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

MFinn9
 one year 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.

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

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