Confusion about derivative of $\arctan(x)+\arctan(\frac{1}{x})$

Question

I was always taught that if a function's derivative is always $0$ on an interval $(a,b)$ then the function is constant on $(a,b)$. And all the information online points to that conclusion as well - in fact, it's proved here!

Yet, when I mentioned it an a conversation with the calculus teacher at our school, he mysteriously told me to look at the function: $$f(x) = \arctan(x)+\arctan(\frac{1}{x})$$

And looking at it, it indeed seems to contradict this theorem. A simple application of the chain rule shows $f'(x)=0$, yet, $f(1)\neq f(-1)$.

Is this apparent contradiction a flaw in the "constant derivative theorem" or am I missing something? How could one amend the statement of the theorem to repair such contradictions?

Answer 1

The domain of $f$ is not connected, $f$ is constant on each connected component.

Answer 2

The theorem is OK, it is your interval that doesn't work. The function is not defined in $0$. So, the function will be constant on $(-\infty, 0)$ and on $(0,+\infty)$, but those two constants need not match.

Answer 3

Hint:

Using Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?,

$$\arctan x+\arctan\dfrac1x=\begin{cases}\arctan x+\text{arccot}x=\dfrac\pi2 &\mbox{if }x\ge0 \\\arctan x+\text{arccot}x-\pi=? & \mbox{if }x<0\end{cases} $$