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

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

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Other answers:

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..
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
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?
\[x + \tan^{-1}x^2 + \tan^{-1}x = 0\] \[x^2.\tan^{-1}x + x + \tan^{-1}x = 0\] \[D = b^2 - 4ac\] \[D = 1 - 4\tan^{-2}x\] \[x = \frac{-1 \pm \sqrt{1-4\tan^{-2}x}}{2\tan^{-1}x}\]
Well, I am just trying.. Ha ha ha..
:O i dont understand anyway can u check if this is right...two different methods gave me two different answers :(
|dw:1341311735216:dw| only when x tends to 0 when when there is 1/x instead of x it tends to infinity
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 :)
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
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
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
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
i mean "and so your derivative is increasing for all \(x\)"
i got it..thnx :D

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