I am trying to understand Daniel Fischer's answer to this question Antiderivative of $\sin(\frac{1}{x})$
I understand the ideas and every step except for the part where he said
The first part $x^2\cos(x^{−1})$ is easily seen to be differentiable at $0$ with derivative $0$ there
But isn't $x^2\cos(x^{−1})$ not defined at $x=0$? So how can we talk about its derivative there? I can see that the limit at $0$ exists, but it's still not defined there. Is he using some properties of $F'$ that I am not seeing? Or do we actually have somehow have this point defined to be zero at zero? Since by Show that the function $g(x) = x^2 \sin(\frac{1}{x}) ,(g(0) = 0)$ is everywhere differentiable and that $g′(0) = 0$ we know that the derivative is zero if we define $x^2\cos(x^{−1})$ to take the value zero at zero, so do we have this? Is there something about how we define $F$ and/or $F'$ that forces this to be true?
I can't really be sure of this, any help is appreciated.