I calculated the derivative of $\arctan\left(\frac{1+x}{1-x}\right)$ to be $\frac{1}{1+x^2}$. This is the same as $(\arctan)'$. Why is there no $c$ that satisfies $\arctan\left(\frac{1+x}{1-x}\right) = \arctan(x) +c$?
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2$\begingroup$ Why do you say that there is no $c$? $\endgroup$ – Thomas Andrews Apr 8 '15 at 2:21
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1$\begingroup$ This seems like one of those cases where trig identities involving constants are not obvious... $\endgroup$ – TravisJ Apr 8 '15 at 2:22
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5$\begingroup$ Plotting the two clearly shows they don't differ by a constant but also provides a hint as to why not. $\endgroup$ – Cascabel Apr 8 '15 at 18:23
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$\begingroup$ Piecewise, they do differ by a constant. Just with a discontinuity at x=1. And we explicitly define arctan(t) to have a discontinuity at t=+/-inf $\endgroup$ – smci Apr 9 '15 at 11:38
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The problem is that $\arctan \frac{1+x}{1-x}$ isn't defined at $x = 1$, and in particular isn't differentiable there. In fact, we have $$ \arctan \frac{1+x}{1-x} - \arctan x = \begin{cases} \frac{\pi}{4} & x < 1, \\ -\frac{3\pi}{4} & x > 1. \end{cases} $$ So the difference is piecewise constant.
answered Apr 8 '15 at 2:23
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$\begingroup$ While formally you are right in the first sentence, note that $\frac x{x^2}$ is not defined at $0$, but it has a continuation there, which is also differentiable at $0$. Here the discontinuity is not a removable one, which is why this doesn't work. $\endgroup$ – Asaf Karagila♦ Apr 8 '15 at 11:23
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1$\begingroup$ @AsafKaragila: I don't think you are talking about the function which you meant to talk about. The function you mentioned cannot be extended continuously at $x=0$. $\endgroup$ – Marc van Leeuwen Apr 8 '15 at 13:06
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1$\begingroup$ @Marc: You're right, I wrote that comment from my phone, so MathJax mistakes are easy to come by. And of course, that I meant $\frac{x^2}{x}$. Thanks for pointing that out! $\endgroup$ – Asaf Karagila♦ Apr 8 '15 at 13:13
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$\begingroup$ d00d, what a kewl constant! It's like not even constant! $\endgroup$ – bjb568 Apr 8 '15 at 21:19
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Hint: $$\tan(x+y)=\frac{\tan x + \tan y}{1-\tan x\tan y}$$
What happens when $\tan y=1$?
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$\begingroup$ The problem is that $\arctan$ always returns an answer in $(-\pi/2,\pi/2)$. $\endgroup$ – Yuval Filmus Apr 8 '15 at 2:24
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Piecewise, they do differ by a constant. Just with a discontinuity at x=1.
And we explicitly define arctan(t) to have that discontinuity at t=+/-inf
