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angela210793

  • 3 years ago

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

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  1. lgbasallote
    • 3 years ago
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    i believe the first step would be to differentiate =_= this sounds complicated...

  2. matricked
    • 3 years ago
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    is it y' =0 when x=0

  3. waterineyes
    • 3 years ago
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    \[\frac{d(x*\tan^{-1}x)}{dx} = \frac{x}{1 + x^2} + \tan^{-1}x\]

  4. lgbasallote
    • 3 years ago
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    i wonder how we can solve for x \[\large \frac{x}{1+ x^2} + \tan^{-1} x = 0\]

  5. mayank_mak
    • 3 years ago
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    yup it is when x = 0 and no other solution

  6. lgbasallote
    • 3 years ago
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    we all know that..but *how* do we prove it..

  7. lgbasallote
    • 3 years ago
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    this is hard to prove i think..

  8. mayank_mak
    • 3 years ago
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    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. lgbasallote
    • 3 years ago
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    hmm i wonder if it's possible algebraically

  10. angela210793
    • 3 years ago
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    i have to study the monotony ,asimptotes and then graph it-_- and i haven't found the roots yet :(

  11. mayank_mak
    • 3 years ago
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    no not possible that way @igbasallote

  12. mayank_mak
    • 3 years ago
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    yes @angela210793 u have to study that to solve these type of questions.

  13. angela210793
    • 3 years ago
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    i know...but i can't find the roots :'(

  14. mayank_mak
    • 3 years ago
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    so better study these topics first then find the roots

  15. angela210793
    • 3 years ago
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    oh wait...(atctg)'=1/1+x^2 no?

  16. waterineyes
    • 3 years ago
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    Yes you are doing wrong earlier..

  17. angela210793
    • 3 years ago
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    when y'=o?

  18. waterineyes
    • 3 years ago
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    \[x + \tan^{-1}x^2 + \tan^{-1}x = 0\] \[x^2.\tan^{-1}x + x + \tan^{-1}x = 0\] \[D = b^2 - 4ac\] \[D = 1 - 4\tan^{-2}x\] \[x = \frac{-1 \pm \sqrt{1-4\tan^{-2}x}}{2\tan^{-1}x}\]

  19. waterineyes
    • 3 years ago
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    Well, I am just trying.. Ha ha ha..

  20. angela210793
    • 3 years ago
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    :O i dont understand anyway can u check if this is right...two different methods gave me two different answers :(

  21. angela210793
    • 3 years ago
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    |dw:1341311440181:dw|

  22. mayank_mak
    • 3 years ago
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    |dw:1341311735216:dw| only when x tends to 0 when when there is 1/x instead of x it tends to infinity

  23. mayank_mak
    • 3 years ago
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    answer to previous question . ps - ignore my bad writing.

  24. agentx5
    • 3 years ago
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    Your handwriting is fine @mayank_mak , gj :-D

  25. angela210793
    • 3 years ago
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    @mayank_mak thanks a lot :D and ur handwriting is very nice :D

  26. anonymous
    • 3 years ago
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    we can prove it if you like

  27. angela210793
    • 3 years ago
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    yes please :)

  28. anonymous
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    derivative of wht is 2/(x^2+1)^2???

  31. anonymous
    • 3 years ago
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    derivative of \[\frac{x}{x^2+1}+\tan^{-1}(x)\]

  32. angela210793
    • 3 years ago
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    y did u find derivative of tht??

  33. anonymous
    • 3 years ago
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    to prove that it was strictly increasing

  34. angela210793
    • 3 years ago
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    ohh..i got tht...

  35. angela210793
    • 3 years ago
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    i thought u were still explaining monotony

  36. anonymous
    • 3 years ago
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    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
    • 3 years ago
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    i mean "and so your derivative is increasing for all \(x\)"

  38. angela210793
    • 3 years ago
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    i got it..thnx :D

  39. anonymous
    • 3 years ago
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    yw

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