$$\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$$
The lecturer made the comment "the domain of x is between -1 and 1". if the domain of x is 1, the equation turns into
$$\frac{1}{\sqrt{1-1^2}}$$ and x becomes 0
$$\frac{1}{0}$$
Why is 1 a domain of x since it will cause the equation to divide by zero which is indeterminate
The step is made for the proof that
$$(\arcsin x)'=\frac{1}{\sqrt{1-x^2}}$$
which is indeed not defined for $x=\pm 1$, that is $x\in(-1,1)$, since the tangent at that point is vertical.