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I'm teaching baby calculus recitation this semester, and I meet a problem to calculate the derivative of $$y=\tan^{-1}(x-\sqrt{1+x^2})$$ Just apply the chain rule and after some preliminary algebra, I find $$\frac{dy}{dx}=\frac{1}{2(1+x^2)}$$ What surprises me is that the result implies $$y=\frac{1}{2}\tan^{-1}x+C$$ Can anyone tell me how to see that directly?

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    $\begingroup$ This is a good finding. Seriously. $\endgroup$
    – NoChance
    Commented Oct 6, 2015 at 2:34
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    $\begingroup$ I might not have posted an answer to a question so many had already answered, but for the fact that I think I've shown a simpler way to do it: the “cartographer's” tangent half-angle formula $\tan\theta\pm\sec\theta = \tan\left( \frac\pi4 \pm \frac\theta2 \right)$ with $x=\tan\theta$ and $\sqrt{1+x^2} = \sec\theta$. See below. ${}\qquad{}$ $\endgroup$ Commented Oct 6, 2015 at 18:39

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Consider the following figure:

enter image description here

Triangles $\triangle ABC$ and $\triangle ACD$ are right triangles, and $AB\cong BD$, so $\triangle ABD$ is isosceles with

$$\angle BAD=\angle BDA={1\over2}(\pi-\angle ABD)={1\over2}\left(\pi-\left({\pi\over2}-\angle BAC \right)\right)={\pi\over4}+{1\over2}\angle BAC$$

where all angles are taken in a positive sense. We also have

$$\angle CAD=\angle BAD-\angle BAC={\pi\over4}-{1\over2}\angle BAC$$

Finally, taking care with minus signs, we have

$$\angle CAD=-\tan^{-1}(x-\sqrt{1+x^2})\quad\text{and}\quad\angle BAC=\tan^{-1}x$$

which gives the desired identity

$$\tan^{-1}(x-\sqrt{1+x^2})={1\over2}\tan^{-1}x-{\pi\over4}$$

Remark (added later): The picture and proof here assume $x\ge0$. To cover the case $x\lt0$ requires either its own picture or an appeal to some kind of analytic continuity.

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  • $\begingroup$ Very nice. For $x<0$, you can instead use $x+\sqrt{1+x^2}$, which places $D$ above $B$, but you still have an isoceles triangle ADB, and with mostly the same derivation, you get $\arctan(x+\sqrt{1+x^2})=\frac12\arctan(x)+\pi/4$. Geometry is beautiful :) $\endgroup$ Commented Oct 6, 2015 at 15:27
  • $\begingroup$ Beautiful proof. Thank you very much! $\endgroup$
    – student
    Commented Oct 6, 2015 at 19:48
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By virtue of solving for $x$, the original equation $$y = \tan^{-1} (x - \sqrt{1+x^2})$$ implies $$x = \frac{1}{2}(\tan y - \cot y) = \frac{\sin^2 y - \cos^2 y}{2 \sin y \cos y} = -\cot 2y,$$ hence $$y = \frac{1}{2}\cot^{-1} (-x) = -\frac{1}{2} \cot^{-1} x.$$ Now recalling that $$\cot^{-1} x + \tan^{-1} x = \frac{\pi}{2},$$ we readily obtain $$y = \frac{1}{2} \tan^{-1} x - \frac{\pi}{4}.$$

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  • $\begingroup$ From $x=-\cot 2y$, how do you conclude that $y=-\frac12\cot^{-1} x$? I mean, since cotangent is $\pi$-periodic, you know for sure that $y=-\frac12\cot^{-1} x+k\pi/2$, but how do you get rid of this $k$? You may use the fact that $y\in ]-\pi/2,0[$, anyway I think there is something missing in your proof. $\endgroup$ Commented Oct 6, 2015 at 11:13
  • $\begingroup$ Also, your identity $\cot^{-1}x+\tan^{-1} x=\pi/2$ is wrong. $\endgroup$ Commented Oct 6, 2015 at 11:24
  • $\begingroup$ Very neat proof! Thank you very much. $\endgroup$
    – student
    Commented Oct 6, 2015 at 19:47
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Let me rewrite the proof a bit differently.

We are required to prove (RTP) that $$2\tan^{-1}(y - \sqrt{1+y^2}) - \tan^{-1} (y) = -\frac{\pi}{2}$$

Proof:

Plan: We will use the formula F first, followed by the identity I.

Let us call the following identity I:

$\tan^{-1}(x) + \tan^{-1} (1/x) =$ \begin{cases} \hfill \pi/2 & \text{if $x\geq 0$}\\ -\pi/2 & \text{if $x<0$} \end{cases}

Next, call the following formula F:

$2 \tan^{-1} (x) =$ \begin{cases} \hfill \tan^{-1} \frac{2x}{1-x^2} & \text{if $|x|\neq 1$}\\ \frac{\pi}{2} & \text{if $x = 1$}\\ -\frac{\pi}{2} & \text{if $x = -1$} \end{cases}

Also note that $x - \sqrt{1+x^2} < 0$, therefore $|x - \sqrt{1+x^2}| > 1$ is equivalent to $x - \sqrt{1+x^2} < -1 \implies x < 0$

Done with the ground work!

If $y < 0$, LHS of RTP =

$$2\tan^{-1}(y - \sqrt{1+y^2}) - \tan^{-1} (y) = -\tan^{-1}\frac{1}{y} - \tan^{-1}(y) = -\frac{\pi}{2}$$

[we used formula F (first case) first, followed by identity I (second case)]

If $y > 0$, LHS of RTP =

$$2\tan^{-1}(y - \sqrt{1+y^2}) - \tan^{-1} (y) = -\tan^{-1}\frac{1}{y} - \tan^{-1}(y) = -\frac{\pi}{2}$$

[we used formula F (first case) first, followed by identity I (first case)]

The case $y = 0$ is obvious!

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    $\begingroup$ This is quite literally the same methodology as what OP did... $\endgroup$ Commented Oct 6, 2015 at 2:58
  • $\begingroup$ @ Cameron Williams: Hope I could clarify what I meant! $\endgroup$
    – PTDS
    Commented Oct 6, 2015 at 3:43
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    $\begingroup$ You still have to show that $2\arctan(x-\sqrt{1+x^2}) = -\arctan\frac{1}{x}$. It isn't obvious to me that this is true. $\endgroup$ Commented Oct 6, 2015 at 3:45
  • $\begingroup$ @ Cameron Williams: Is it okay now? $\endgroup$
    – PTDS
    Commented Oct 6, 2015 at 3:52
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    $\begingroup$ I think so. This is a better answer than mine, I think. Take the down votes to mean that you should explain your answers better in the future. It seems like you know what you're saying, but it can be difficult to tell what you mean if you don't explain well to us. $\endgroup$ Commented Oct 6, 2015 at 3:56
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First let's figure out what $C$ is. Letting $x=0$, we see that $\arctan(-1) = C$ (i.e. $C = -\frac{\pi}{4}$ or $C = \frac{3\pi}{4}$). Since inverse trigonometric functions are hard to work with, let's take $\tan$ of both sides. Doing so we get

$$\tan\left(\arctan(x-\sqrt{1+x^2})\right) = \tan\left(\frac{1}{2}\arctan(x)+C\right).$$

The left hand side is nothing more than $x-\sqrt{1+x^2}$. As for the right hand side, we can employ the tangent sum formula:

$$\tan(A+B) = \frac{\tan A + \tan B}{1-\tan A\tan B}.$$

Thus

$$\tan\left(\frac{1}{2}\arctan(x)+C\right) = \frac{\tan\left(\frac{1}{2}\arctan(x)\right) + \tan(\arctan(-1))}{1-\tan\left(\frac{1}{2}\arctan(x)\right)\tan(\arctan(-1))}.$$

We have that $\tan\left(\frac{1}{2}\arctan(x)\right) = \frac{\sqrt{1+x^2}-1}{x}$ which can be verified by half angle formulas. If you put these pieces together and do some algebra, you will have your answer.

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  • $\begingroup$ Well, I cannot deny it's correct, but I hope there is a neat way to see it. I'll just upvote for it, and I'll check it if no other simpler answer occur later. Anyway, thanks for your answer! $\endgroup$
    – student
    Commented Oct 6, 2015 at 2:27
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    $\begingroup$ Yeah this approach lacks a certain finesse, but I'm not sure how much more elegant it can be without appealing to to this process in some way, shape, or form. I'd be really interested to see if anyone has a better approach. $\endgroup$ Commented Oct 6, 2015 at 2:29
  • $\begingroup$ To prove two numbers are equal, you prove their tangents are equal. Fundamentally wrong. $\endgroup$ Commented Oct 6, 2015 at 11:04
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Let $x=\tan\theta$. Then what I like to think of as the “cartographer's” tangent half-angle formula says $$ \tan\theta \pm \sec\theta = \tan\left(\frac\pi4 \pm \frac \theta 2\right). $$

If $\sec\theta>0$ then $\sec\theta=\sqrt{1+x^2}$.

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First, you have $\arctan (\tan y)=y$ for $y\in]-\pi/2, \pi/2[$.

And for all $y\neq \pi/4+k\pi/2$, $$\tan (2y)=\frac{2\tan y}{1-\tan^2 y}$$

Thus, for all $z\in]-1,1[$, $y=\arctan z$ is in $]-\pi/4,\pi/4[$, and you can apply the above:

$$\tan(2\arctan z)=\frac{2z}{1-z^2}$$

And since $2\arctan z \in ]-\pi/2,\pi/2[$, you can also apply the first identity to get

$$\arctan \tan (2\arctan z)=2\arctan z=\arctan \left(\frac{2z}{1-z^2}\right)$$

Bear in mind that this is proved only for $z\in]-1,1[$, and it actually is wrong outside this interval.

Now, for $x>0$, $x-\sqrt{1+x^2}<0$ since $1+x^2>x^2$ implies $\sqrt{1+x^2}>x$.

But it's also $>-1$, since

$$(1+x)^2=x^2+1+2x>x^2+1$$

And taking square root, $1+x>\sqrt{1+x^2}$

Thus, for $x>0$, $-1<x-\sqrt{1+x^2}<0$.

It's thus a good candidate to apply the formula above with arctangent, which is true for $|z|<1$:

That is, for $x>0$,

$$2\arctan (x-\sqrt{1+x^2})=\arctan \left(\frac{2(x-\sqrt{1+x^2})}{1-(x-\sqrt{1+x^2})^2}\right)$$

And

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

Thus, still for $x>0$,

$$2\arctan (x-\sqrt{1+x^2})=-\arctan\frac{1}{x}$$

And since $\arctan x+\arctan \frac1x=\pi/2$ for $x>0$, you have

$$2\arctan (x-\sqrt{1+x^2})=\arctan (x) -\pi/2$$


Now, we have to prove the same for $x<0$.

Then, $x-\sqrt{1+x^2}<-1$, so we can't apply our formula with arctangent. However, its inverse is in $]-1,0[$, and for $x<0$,

$$\arctan (x-\sqrt{1+x^2})+\arctan \left(\frac{1}{x-\sqrt{1+x^2}}\right)=-\pi/2$$

$$\arctan (x-\sqrt{1+x^2})-\arctan (x+\sqrt{1+x^2})=-\pi/2$$ $$\arctan (x-\sqrt{1+x^2})=\arctan (x+\sqrt{1+x^2})-\pi/2$$

Then, still for $x<0$, $x+\sqrt{1+x^2} \in ]0,1[$ and

$$2\arctan (x+\sqrt{1+x^2})=\arctan \frac{2(x+\sqrt{1+x^2})}{1-(x+\sqrt{1+x^2})^2}$$

$$2\arctan (x+\sqrt{1+x^2})=\arctan \frac{2(x+\sqrt{1+x^2})}{-2x(x+\sqrt{1+x^2})}=-\arctan \frac1x$$

And $\arctan x+\arctan \frac1x=-\pi/2$ (remember that $x<0$), hence

$$2\arctan (x+\sqrt{1+x^2})=\arctan (x)+\pi/2$$

And finally

$$2\arctan (x-\sqrt{1+x^2})=2\arctan (x+\sqrt{1+x^2})-\pi=\arctan (x)-\pi/2$$

This formula is now proved for $x>0$ and for $x<0$. It's also true for $x=0$, since then it amounts to $2\arctan (-1)=-\pi/2$.

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