The question is:
Find the derivative of $\arctan\left(\frac {1} {x}\right)$ and specify for which $x$ the derivative exists
I get that $f'(x) = - \dfrac{1}{x^2+1}$ and for that, I say that the derivative exists for all real numbers since the denominator cannot be equal to $0$. However, the answer says that the derivative exists for all numbers except $x = 0$. Why is that?
Realize that $\arctan\left({1\over x}\right)$ is not continuous at $x=0$, thus not differentiable. This is because $\arctan\left({1\over0}\right)$ does not exist and:
$$\lim_{x\to 0^-}\arctan\left({1\over x}\right) \neq \lim_{x\to 0^+}\arctan\left({1\over x}\right)$$
It may seem like $\arctan\left({1\over0}\right)$ should be $\pi \over 2$ because $\arctan(\infty) = {\pi\over2}$, but since:
$$\lim_{x\to +\infty} \arctan x\neq \lim_{x\to -\infty} \arctan x$$
You cannot make such a conclusion. The domain of $\arctan\left({1\over x}\right)$ is $\Bbb R \setminus \{0\}$ so discontinuity implies indifferentiability.
Note that without simplifying the derivative, you'll get a hole at $x = 0$:
$$\frac d {dx} \arctan \left({1\over x}\right) = {1 \over \left(\color{red}{1\over x}\right)^2 + 1} \cdot -{1 \over x^2}$$
Similar to how you can factor and cancel in rational functions to get rid of a hole and evaluate an otherwise undefined limit:
$$\require{cancel} \lim_{x \to 4} {\cancel{(x-4)}(x-5) \over \cancel{x-4}} = -1$$
Even though simplifying the derivative gives you $-\frac 1{x^2 + 1}$, there is still a hole in the unsimplified derivative—meaning the original function's not differentiable there.