## angela210793 4 years ago Y=x*arctanx when does y'=0???

1. anonymous

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

2. anonymous

is it y' =0 when x=0

3. anonymous

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

4. anonymous

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

5. anonymous

yup it is when x = 0 and no other solution

6. anonymous

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

7. anonymous

this is hard to prove i think..

8. 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

9. anonymous

hmm i wonder if it's possible algebraically

10. angela210793

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

11. anonymous

no not possible that way @igbasallote

12. anonymous

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

13. angela210793

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

14. anonymous

so better study these topics first then find the roots

15. angela210793

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

16. anonymous

Yes you are doing wrong earlier..

17. angela210793

when y'=o?

18. anonymous

$x + \tan^{-1}x^2 + \tan^{-1}x = 0$ $x^2.\tan^{-1}x + x + \tan^{-1}x = 0$ $D = b^2 - 4ac$ $D = 1 - 4\tan^{-2}x$ $x = \frac{-1 \pm \sqrt{1-4\tan^{-2}x}}{2\tan^{-1}x}$

19. anonymous

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

20. angela210793

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

21. angela210793

|dw:1341311440181:dw|

22. anonymous

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

23. anonymous

24. anonymous

Your handwriting is fine @mayank_mak , gj :-D

25. angela210793

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

26. anonymous

we can prove it if you like

27. angela210793

28. 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

29. 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

30. angela210793

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

31. anonymous

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

32. angela210793

y did u find derivative of tht??

33. anonymous

to prove that it was strictly increasing

34. angela210793

ohh..i got tht...

35. angela210793

i thought u were still explaining monotony

36. 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

37. anonymous

i mean "and so your derivative is increasing for all $$x$$"

38. angela210793

i got it..thnx :D

39. anonymous

yw