Y=x*arctanx
when does y'=0???

- angela210793

Y=x*arctanx
when does y'=0???

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- lgbasallote

i believe the first step would be to differentiate =_= this sounds complicated...

- anonymous

is it y' =0 when x=0

- anonymous

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

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## More answers

- lgbasallote

i wonder how we can solve for x
\[\large \frac{x}{1+ x^2} + \tan^{-1} x = 0\]

- anonymous

yup it is when x = 0 and no other solution

- lgbasallote

we all know that..but *how* do we prove it..

- lgbasallote

this is hard to prove i think..

- 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

hmm i wonder if it's possible algebraically

- angela210793

i have to study the monotony ,asimptotes and then graph it-_- and i haven't found the roots yet :(

- anonymous

no not possible that way @igbasallote

- anonymous

yes @angela210793 u have to study that to solve these type of questions.

- angela210793

i know...but i can't find the roots :'(

- anonymous

so better study these topics first then find the roots

- angela210793

oh wait...(atctg)'=1/1+x^2 no?

- anonymous

Yes you are doing wrong earlier..

- angela210793

when y'=o?

- 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

Well, I am just trying..
Ha ha ha..

- angela210793

:O i dont understand anyway can u check if this is right...two different methods gave me two different answers :(

- angela210793

|dw:1341311440181:dw|

- anonymous

|dw:1341311735216:dw| only when x tends to 0
when when there is 1/x instead of x it tends to infinity

- anonymous

answer to previous question .
ps - ignore my bad writing.

##### 2 Attachments

- anonymous

Your handwriting is fine @mayank_mak , gj :-D

- angela210793

@mayank_mak thanks a lot :D
and ur handwriting is very nice :D

- anonymous

we can prove it if you like

- angela210793

yes please :)

- 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

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

derivative of wht is 2/(x^2+1)^2???

- anonymous

derivative of
\[\frac{x}{x^2+1}+\tan^{-1}(x)\]

- angela210793

y did u find derivative of tht??

- anonymous

to prove that it was strictly increasing

- angela210793

ohh..i got tht...

- angela210793

i thought u were still explaining monotony

- 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

i mean "and so your derivative is increasing for all \(x\)"

- angela210793

i got it..thnx :D

- anonymous

yw

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