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lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
i believe the first step would be to differentiate =_= this sounds complicated...
 2 years ago

matricked Group TitleBest ResponseYou've already chosen the best response.1
is it y' =0 when x=0
 2 years ago

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

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

mayank_mak Group TitleBest ResponseYou've already chosen the best response.3
yup it is when x = 0 and no other solution
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
we all know that..but *how* do we prove it..
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
this is hard to prove i think..
 2 years ago

mayank_mak Group TitleBest 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
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
hmm i wonder if it's possible algebraically
 2 years ago

angela210793 Group TitleBest 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 :(
 2 years ago

mayank_mak Group TitleBest ResponseYou've already chosen the best response.3
no not possible that way @igbasallote
 2 years ago

mayank_mak Group TitleBest ResponseYou've already chosen the best response.3
yes @angela210793 u have to study that to solve these type of questions.
 2 years ago

angela210793 Group TitleBest ResponseYou've already chosen the best response.0
i know...but i can't find the roots :'(
 2 years ago

mayank_mak Group TitleBest ResponseYou've already chosen the best response.3
so better study these topics first then find the roots
 2 years ago

angela210793 Group TitleBest ResponseYou've already chosen the best response.0
oh wait...(atctg)'=1/1+x^2 no?
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.0
Yes you are doing wrong earlier..
 2 years ago

angela210793 Group TitleBest ResponseYou've already chosen the best response.0
when y'=o?
 2 years ago

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

waterineyes Group TitleBest ResponseYou've already chosen the best response.0
Well, I am just trying.. Ha ha ha..
 2 years ago

angela210793 Group TitleBest 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 :(
 2 years ago

angela210793 Group TitleBest ResponseYou've already chosen the best response.0
dw:1341311440181:dw
 2 years ago

mayank_mak Group TitleBest 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
 2 years ago

mayank_mak Group TitleBest ResponseYou've already chosen the best response.3
answer to previous question . ps  ignore my bad writing.
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
Your handwriting is fine @mayank_mak , gj :D
 2 years ago

angela210793 Group TitleBest ResponseYou've already chosen the best response.0
@mayank_mak thanks a lot :D and ur handwriting is very nice :D
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
we can prove it if you like
 2 years ago

angela210793 Group TitleBest ResponseYou've already chosen the best response.0
yes please :)
 2 years ago

satellite73 Group TitleBest 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
 2 years ago

satellite73 Group TitleBest 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
 2 years ago

angela210793 Group TitleBest ResponseYou've already chosen the best response.0
derivative of wht is 2/(x^2+1)^2???
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
derivative of \[\frac{x}{x^2+1}+\tan^{1}(x)\]
 2 years ago

angela210793 Group TitleBest ResponseYou've already chosen the best response.0
y did u find derivative of tht??
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
to prove that it was strictly increasing
 2 years ago

angela210793 Group TitleBest ResponseYou've already chosen the best response.0
ohh..i got tht...
 2 years ago

angela210793 Group TitleBest ResponseYou've already chosen the best response.0
i thought u were still explaining monotony
 2 years ago

satellite73 Group TitleBest 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
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
i mean "and so your derivative is increasing for all \(x\)"
 2 years ago

angela210793 Group TitleBest ResponseYou've already chosen the best response.0
i got it..thnx :D
 2 years ago
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