A community for students.
Here's the question you clicked on:
 0 viewing
angela210793
 4 years ago
Y=x*arctanx
when does y'=0???
angela210793
 4 years ago
Y=x*arctanx when does y'=0???

This Question is Closed

lgbasallote
 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\[\frac{d(x*\tan^{1}x)}{dx} = \frac{x}{1 + x^2} + \tan^{1}x\]

lgbasallote
 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

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

lgbasallote
 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

lgbasallote
 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 \(x\)"
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.