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I don't understand this property of arctan(x)'

  1. May 16, 2009 #1
    The derivative of arctan(x) is [itex]\frac{1}{x^2+1}[/itex]

    The derivative of arctan(x - sqrt(x^2 + 1) is [itex]\frac{1}{2(x^2+1)}[/itex]

    take arctan(x) to be A and arctan(x-sqrt(x^2 + 1)) to be B.


    Why doesn't [itex]\frac{1}{2}*A=B[/itex]?

    If you'd like, I can prove the derivative for you, but it's moderately tedious to simplify it to that point.
  2. jcsd
  3. May 16, 2009 #2


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    Yes, I would like to see your derivation of B'. I get nothing like that!
  4. May 17, 2009 #3
  5. May 17, 2009 #4


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    That's what I came out with!
  6. May 17, 2009 #5


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    I plugged this into mathematica and it indeed simplifies to [tex]\frac{1}{2+2x^2}[/tex]. Which would suggest that [itex]\arctan(x)=2 \arctan(x-\sqrt{x^2+1})+C[/itex]. Therefore [itex] C=\arctan(x)-2\arctan(x-\sqrt{x^2+1})=\frac{\pi}{4}[/itex].

    http://img32.imageshack.us/img32/3028/arctan.jpg [Broken]

    I am not sure how to find C yet, mathematica wasn't able to find it and I am not in the mood to try and symplify it currently. However all the numerical values I tried yielded pi/4.
    Last edited by a moderator: May 4, 2017
  7. May 17, 2009 #6

    George Jones

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    One way is to set

    z=\tan^{-1} \left(x - \sqrt{1 + x^2} \right),

    so that

    [tex]\tan z = x - \sqrt{1 + x^2}[/tex]

    Solving for [itex]x[/itex] then leads directly to your result.
  8. May 17, 2009 #7
    Haven't had a chance to check the thread until now.

    I completely forgot that there's a constant!

    That makes so much sense, thank you everyone.

    P.S. I'm only in the first calc class, this just really bothered me, and I didn't want to wait till class on tuesday to find out why it worked out the way it did.
  9. May 17, 2009 #8
  10. May 18, 2009 #9


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    That's a clever way to solve it George.

    Mr_garlic, your first result was correct. After simplification that ugly looking expression will reduce to the nice expression you found in your first post.
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