angela210793
  • angela210793
Y=x*arctanx when does y'=0???
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

lgbasallote
  • lgbasallote
i believe the first step would be to differentiate =_= this sounds complicated...
anonymous
  • anonymous
is it y' =0 when x=0
anonymous
  • anonymous
\[\frac{d(x*\tan^{-1}x)}{dx} = \frac{x}{1 + x^2} + \tan^{-1}x\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

lgbasallote
  • lgbasallote
i wonder how we can solve for x \[\large \frac{x}{1+ x^2} + \tan^{-1} x = 0\]
anonymous
  • anonymous
yup it is when x = 0 and no other solution
lgbasallote
  • lgbasallote
we all know that..but *how* do we prove it..
lgbasallote
  • lgbasallote
this is hard to prove i think..
anonymous
  • anonymous
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
  • lgbasallote
hmm i wonder if it's possible algebraically
angela210793
  • angela210793
i have to study the monotony ,asimptotes and then graph it-_- and i haven't found the roots yet :(
anonymous
  • anonymous
no not possible that way @igbasallote
anonymous
  • anonymous
yes @angela210793 u have to study that to solve these type of questions.
angela210793
  • angela210793
i know...but i can't find the roots :'(
anonymous
  • anonymous
so better study these topics first then find the roots
angela210793
  • angela210793
oh wait...(atctg)'=1/1+x^2 no?
anonymous
  • anonymous
Yes you are doing wrong earlier..
angela210793
  • angela210793
when y'=o?
anonymous
  • anonymous
\[x + \tan^{-1}x^2 + \tan^{-1}x = 0\] \[x^2.\tan^{-1}x + x + \tan^{-1}x = 0\] \[D = b^2 - 4ac\] \[D = 1 - 4\tan^{-2}x\] \[x = \frac{-1 \pm \sqrt{1-4\tan^{-2}x}}{2\tan^{-1}x}\]
anonymous
  • anonymous
Well, I am just trying.. Ha ha ha..
angela210793
  • angela210793
:O i dont understand anyway can u check if this is right...two different methods gave me two different answers :(
angela210793
  • angela210793
|dw:1341311440181:dw|
anonymous
  • anonymous
|dw:1341311735216:dw| only when x tends to 0 when when there is 1/x instead of x it tends to infinity
anonymous
  • anonymous
answer to previous question . ps - ignore my bad writing.
anonymous
  • anonymous
Your handwriting is fine @mayank_mak , gj :-D
angela210793
  • angela210793
@mayank_mak thanks a lot :D and ur handwriting is very nice :D
anonymous
  • anonymous
we can prove it if you like
angela210793
  • angela210793
yes please :)
anonymous
  • anonymous
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
  • anonymous
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
  • angela210793
derivative of wht is 2/(x^2+1)^2???
anonymous
  • anonymous
derivative of \[\frac{x}{x^2+1}+\tan^{-1}(x)\]
angela210793
  • angela210793
y did u find derivative of tht??
anonymous
  • anonymous
to prove that it was strictly increasing
angela210793
  • angela210793
ohh..i got tht...
angela210793
  • angela210793
i thought u were still explaining monotony
anonymous
  • anonymous
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
  • anonymous
i mean "and so your derivative is increasing for all \(x\)"
angela210793
  • angela210793
i got it..thnx :D
anonymous
  • anonymous
yw

Looking for something else?

Not the answer you are looking for? Search for more explanations.