MFinn9
Derivative of (1+x)^(1/x)
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mathsmind
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have you done something called the D-operator?
mathsmind
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first of all rewrite that as (1+x)^(-x), does that clue or hint help?
mathsmind
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in other words you may use implicit differentiation ...
MFinn9
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so then it's chain rule
mathsmind
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ok just for you to never forget do you know how to find dy/dx of x^x?
MFinn9
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yes using natural logs
mathsmind
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ok then use the same method and differnetiate both sides
MFinn9
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so it would be -1-2x
mathsmind
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impossible to give that answer life is not linear...
MFinn9
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no wait
mathsmind
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y = (1+x)^(-x)
mathsmind
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take the ln of both sides...
mathsmind
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ln(y) = ln[(1+x)^(-x)]
mathsmind
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my appology that the equation editor is not functioning for me tonight for a funny reason...
MFinn9
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ok
mathsmind
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use natural logs rules to obtain: ln(y)=-xln(1+x)
mathsmind
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now can you diff both sides implicitly ...
MFinn9
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f'(x)=-ln(1+x)-[x/(1+x)]*(1+x)^(-x)
mathsmind
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no look at what i wrote again ln(y)=-xln(1+x)
mathsmind
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you used the product rule but u made a small error
MFinn9
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let's work on this step by step
mathsmind
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ok sure
mathsmind
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now the editor is working
mathsmind
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\[ \ln(y)=-xln(1+x)\]
mathsmind
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\[d( \ln(y)=-xln(1+x))\]
MFinn9
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(1/y)y'=-1*ln(x+1)=(-x)(1/(1+x))
mathsmind
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you mean + not =
MFinn9
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yes
mathsmind
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ok all you are left with is y x RHS
mathsmind
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take the y to the other side
mathsmind
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y'=y[-1*ln(x+1)=(-x)(1/(1+x))]
mathsmind
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can you see the y?
MFinn9
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yes... everything** is multiplied by y, not just tacked onto the end
mathsmind
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the next question is what is y=?
mathsmind
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just change the y to the main equation
MFinn9
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now I have to apply that to x -> 0, L'Hopital's Rule
mathsmind
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no
mathsmind
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delete that post
mathsmind
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we don't have limits
mathsmind
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\[y'=[(1+x)^(1/x)][-1*\ln(x+1)=(-x)(1/(1+x))]\]
mathsmind
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can u see i changed the y to (1+x)^(1/x)
MFinn9
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yes, i got that
mathsmind
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good let's us wrap it up
mathsmind
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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
mathsmind
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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...
MFinn9
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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
mathsmind
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that is a different question now
mathsmind
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the answer would be 1/e
mathsmind
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it is easy to see
MFinn9
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Please show me how it is easy to see.
mathsmind
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sorry the answer would be e
mathsmind
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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
mathsmind
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ok to make life esear
mathsmind
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you should use LH rule
mathsmind
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let me prove that for you
MFinn9
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yes, I need to understand this.
mathsmind
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\[y = (1+x)^{(1/x)}\]
mathsmind
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\[\ln(y) =\ln[ (1+x)^{(1/x)}]\]
mathsmind
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\[\ln(y) =\frac{\ln[ (1+x)}{x}\]
Ephilo
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i actually just took a quiz on this
MFinn9
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are you at hopkins by chance? haha... I want to understand this
mathsmind
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\[\ln(y) = \lim_{x \rightarrow 0} [\frac{\ln[ (1+x)}{x}]\]
Ephilo
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i say (1+x)^1/x-1/x
Ephilo
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im prob wrong tho lol
mathsmind
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now differentiate the top and the butoom using LH rule...
mathsmind
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can you see what will happen
MFinn9
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(1/1+x)/x
mathsmind
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\[\ln(y) = \lim_{x \rightarrow 0} [\frac{\frac{1}{[ (1+x)}}{1}]\]
mathsmind
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yes plug in the 0
MFinn9
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oh yes, 1
MFinn9
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sorry
mathsmind
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ln y =1
mathsmind
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y =e
mathsmind
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hehehehe
MFinn9
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I see!
mathsmind
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its a clever qs
mathsmind
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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 ...
mathsmind
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then it will*
mathsmind
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y = e^1
mathsmind
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which means y =e
mathsmind
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did you get that...or am i disconnected...
MFinn9
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no, i see that... much more advanced that what I'm used to seeing, but it makes sense.
mathsmind
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well good luck and all the best
MFinn9
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Thank you!
mathsmind
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you're welcome
mathsmind
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where is my medal hehehe