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

[mr_garlic]

Main Question or Discussion Point

The derivative of arctan(x) is $\frac{1}{x^2+1}$

The derivative of arctan(x - sqrt(x^2 + 1) is $\frac{1}{2(x^2+1)}$

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

$A'=\frac{1}{x^2+1}$
$B'=\frac{1}{2}*A'$

Why doesn't $\frac{1}{2}*A=B$?

If you'd like, I can prove the derivative for you, but it's moderately tedious to simplify it to that point.

 

[HallsofIvy]

The derivative of arctan(x) is $\frac{1}{x^2+1}$

The derivative of arctan(x - sqrt(x^2 + 1) is $\frac{1}{2(x^2+1)}$

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

$A'=\frac{1}{x^2+1}$
$B'=\frac{1}{2}*A'$

Why doesn't $\frac{1}{2}*A=B$?

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!

 

[Irrational]

The derivative of arctan(x - sqrt(x^2 + 1) is $\frac{1}{2(x^2+1)}$

Wolfram solution

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

 

[HallsofIvy]

That's what I came out with!

 

[Cyosis]

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

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.

 

[George Jones]

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

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

so that

$$\tan z = x - \sqrt{1 + x^2}$$

Solving for $x$ then leads directly to your result.

 

[mr_garlic]

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.

 

[mr_garlic]

Wolfram solution

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

Not to double-post, but Wolfram didn't consider the chain rule as of yesterday! That's impressive progress.

 

[Cyosis]

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.