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

is it y' =0 when x=0

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

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

yup it is when x = 0 and no other solution

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

this is hard to prove i think..

hmm i wonder if it's possible algebraically

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

no not possible that way @igbasallote

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

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

so better study these topics first then find the roots

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

Yes you are doing wrong earlier..

when y'=o?

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

|dw:1341311440181:dw|

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

Your handwriting is fine @mayank_mak , gj :-D

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

we can prove it if you like

yes please :)

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

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

y did u find derivative of tht??

to prove that it was strictly increasing

ohh..i got tht...

i thought u were still explaining monotony

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

i got it..thnx :D

yw