angela210793
Y=x*arctanx
when does y'=0???
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lgbasallote
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i believe the first step would be to differentiate =_= this sounds complicated...
matricked
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is it y' =0 when x=0
waterineyes
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\[\frac{d(x*\tan^{-1}x)}{dx} = \frac{x}{1 + x^2} + \tan^{-1}x\]
lgbasallote
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i wonder how we can solve for x
\[\large \frac{x}{1+ x^2} + \tan^{-1} x = 0\]
mayank_mak
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yup it is when x = 0 and no other solution
lgbasallote
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we all know that..but *how* do we prove it..
lgbasallote
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this is hard to prove i think..
mayank_mak
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take tan(inverse)x on other side
and draw graphs of both
arctan(x)
and -x/(1+x^2)
you will find their graphs cutting each other at x = 0 only
lgbasallote
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hmm i wonder if it's possible algebraically
angela210793
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i have to study the monotony ,asimptotes and then graph it-_- and i haven't found the roots yet :(
mayank_mak
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no not possible that way @igbasallote
mayank_mak
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yes @angela210793 u have to study that to solve these type of questions.
angela210793
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i know...but i can't find the roots :'(
mayank_mak
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so better study these topics first then find the roots
angela210793
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oh wait...(atctg)'=1/1+x^2 no?
waterineyes
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Yes you are doing wrong earlier..
angela210793
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when y'=o?
waterineyes
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\[x + \tan^{-1}x^2 + \tan^{-1}x = 0\]
\[x^2.\tan^{-1}x + x + \tan^{-1}x = 0\]
\[D = b^2 - 4ac\]
\[D = 1 - 4\tan^{-2}x\]
\[x = \frac{-1 \pm \sqrt{1-4\tan^{-2}x}}{2\tan^{-1}x}\]
waterineyes
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Well, I am just trying..
Ha ha ha..
angela210793
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:O i dont understand anyway can u check if this is right...two different methods gave me two different answers :(
angela210793
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|dw:1341311440181:dw|
mayank_mak
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|dw:1341311735216:dw| only when x tends to 0
when when there is 1/x instead of x it tends to infinity
mayank_mak
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answer to previous question .
ps - ignore my bad writing.
agentx5
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Your handwriting is fine @mayank_mak , gj :-D
angela210793
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@mayank_mak thanks a lot :D
and ur handwriting is very nice :D
anonymous
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we can prove it if you like
angela210793
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yes please :)
anonymous
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sorry i was writing too much
you have the derivative right? it is
\[\frac{x}{x^2+1}+\tan^{-1}(x)\] and this is clearly zero if \(x=0\) by inspection
anonymous
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not much to that.
now to show that this is the only solution, note that the derivative of this is
\[\frac{2}{(x^2+1)^2}\] and this is always positive, which means your function is strictly increasing
so there is only on solution
angela210793
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derivative of wht is 2/(x^2+1)^2???
anonymous
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derivative of
\[\frac{x}{x^2+1}+\tan^{-1}(x)\]
angela210793
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y did u find derivative of tht??
anonymous
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to prove that it was strictly increasing
angela210793
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ohh..i got tht...
angela210793
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i thought u were still explaining monotony
anonymous
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it is clear that for all \(x\) the function \(\frac{2}{(x+1)^2}>0\) and so it is increasing for all \(x\)
if you have one zero, there cannot be another one
anonymous
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i mean "and so your derivative is increasing for all \(x\)"
angela210793
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i got it..thnx :D
anonymous
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yw