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lgbasalloteBest ResponseYou've already chosen the best response.0
i believe the first step would be to differentiate =_= this sounds complicated...
 one year ago

matrickedBest ResponseYou've already chosen the best response.1
is it y' =0 when x=0
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
\[\frac{d(x*\tan^{1}x)}{dx} = \frac{x}{1 + x^2} + \tan^{1}x\]
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
i wonder how we can solve for x \[\large \frac{x}{1+ x^2} + \tan^{1} x = 0\]
 one year ago

mayank_makBest ResponseYou've already chosen the best response.3
yup it is when x = 0 and no other solution
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
we all know that..but *how* do we prove it..
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
this is hard to prove i think..
 one year ago

mayank_makBest ResponseYou've already chosen the best response.3
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
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
hmm i wonder if it's possible algebraically
 one year ago

angela210793Best ResponseYou've already chosen the best response.0
i have to study the monotony ,asimptotes and then graph it_ and i haven't found the roots yet :(
 one year ago

mayank_makBest ResponseYou've already chosen the best response.3
no not possible that way @igbasallote
 one year ago

mayank_makBest ResponseYou've already chosen the best response.3
yes @angela210793 u have to study that to solve these type of questions.
 one year ago

angela210793Best ResponseYou've already chosen the best response.0
i know...but i can't find the roots :'(
 one year ago

mayank_makBest ResponseYou've already chosen the best response.3
so better study these topics first then find the roots
 one year ago

angela210793Best ResponseYou've already chosen the best response.0
oh wait...(atctg)'=1/1+x^2 no?
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
Yes you are doing wrong earlier..
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
\[x + \tan^{1}x^2 + \tan^{1}x = 0\] \[x^2.\tan^{1}x + x + \tan^{1}x = 0\] \[D = b^2  4ac\] \[D = 1  4\tan^{2}x\] \[x = \frac{1 \pm \sqrt{14\tan^{2}x}}{2\tan^{1}x}\]
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
Well, I am just trying.. Ha ha ha..
 one year ago

angela210793Best ResponseYou've already chosen the best response.0
:O i dont understand anyway can u check if this is right...two different methods gave me two different answers :(
 one year ago

angela210793Best ResponseYou've already chosen the best response.0
dw:1341311440181:dw
 one year ago

mayank_makBest ResponseYou've already chosen the best response.3
dw:1341311735216:dw only when x tends to 0 when when there is 1/x instead of x it tends to infinity
 one year ago

mayank_makBest ResponseYou've already chosen the best response.3
answer to previous question . ps  ignore my bad writing.
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
Your handwriting is fine @mayank_mak , gj :D
 one year ago

angela210793Best ResponseYou've already chosen the best response.0
@mayank_mak thanks a lot :D and ur handwriting is very nice :D
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
we can prove it if you like
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
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
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
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
 one year ago

angela210793Best ResponseYou've already chosen the best response.0
derivative of wht is 2/(x^2+1)^2???
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
derivative of \[\frac{x}{x^2+1}+\tan^{1}(x)\]
 one year ago

angela210793Best ResponseYou've already chosen the best response.0
y did u find derivative of tht??
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
to prove that it was strictly increasing
 one year ago

angela210793Best ResponseYou've already chosen the best response.0
ohh..i got tht...
 one year ago

angela210793Best ResponseYou've already chosen the best response.0
i thought u were still explaining monotony
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
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
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
i mean "and so your derivative is increasing for all \(x\)"
 one year ago

angela210793Best ResponseYou've already chosen the best response.0
i got it..thnx :D
 one year ago
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