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angela210793
 4 years ago
Y=x*arctanx
when does y'=0???
angela210793
 4 years ago
Y=x*arctanx when does y'=0???

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i believe the first step would be to differentiate =_= this sounds complicated...

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yup it is when x = 0 and no other solution

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we all know that..but *how* do we prove it..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is hard to prove i think..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0take 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm i wonder if it's possible algebraically

angela210793
 4 years ago
Best ResponseYou've already chosen the best response.0i have to study the monotony ,asimptotes and then graph it_ and i haven't found the roots yet :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no not possible that way @igbasallote

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes @angela210793 u have to study that to solve these type of questions.

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so better study these topics first then find the roots

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes you are doing wrong earlier..

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, I am just trying.. Ha ha ha..

angela210793
 4 years ago
Best 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 :(

angela210793
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1341311440181:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1341311735216:dw only when x tends to 0 when when there is 1/x instead of x it tends to infinity

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0answer to previous question . ps  ignore my bad writing.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Your handwriting is fine @mayank_mak , gj :D

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we can prove it if you like

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry 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
 4 years ago
Best ResponseYou've already chosen the best response.0not 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
 4 years ago
Best ResponseYou've already chosen the best response.0derivative of wht is 2/(x^2+1)^2???

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0to prove that it was strictly increasing

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it 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
 4 years ago
Best ResponseYou've already chosen the best response.0i mean "and so your derivative is increasing for all \(x\)"
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