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

  • Thread starter mr_garlic
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  • #1
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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.

[itex]A'=\frac{1}{x^2+1}[/itex]
[itex]B'=\frac{1}{2}*A'[/itex]

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.
 

Answers and Replies

  • #2
HallsofIvy
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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.

[itex]A'=\frac{1}{x^2+1}[/itex]
[itex]B'=\frac{1}{2}*A'[/itex]

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.
Yes, I would like to see your derivation of B'. I get nothing like that!
 
  • #4
HallsofIvy
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That's what I came out with!
 
  • #5
Cyosis
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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].

Picture:
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.
 
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  • #6
George Jones
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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.
One way is to set

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

so that

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

Solving for [itex]x[/itex] then leads directly to your result.
 
  • #7
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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
Cyosis
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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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