The function $\arcsin(\sin(x))$ I have this question but I'm really not sure about my solution. Does this make sense?

Given the function $g(x) = \arcsin(\sin x))$, find its domain and where it is differentiable.

So I assume it's enough to say that $\sin$'s co-domain is $[-1,1]$ and so $\arcsin$'s domain, being its inverse function, is $[-1,1]$.
Now $\sin$ is differentiable on $\mathbb R$ and $\arcsin$ is differentiable on $(-1,1)$ (is it?) and so using the chain rule we can say that $g$ is differentiable at all $a \in \mathbb R$ for which $\sin(a) \neq 1$ or $\sin(a) \neq -1$. Which would be $\mathbb R $ \ {$-\pi/2, \pi/2$}.
Thanks!!
 A: A picture is worth a thousand words:

The slopes are $\pm 1$.  Clearly the domain is the whole real line and it is differentiable except at the corners, which come at $\frac \pi2+k\pi$ for $k$ an integer.
A: I'm sorry if this sounds pedantic, but it seems that there are a few things about your post that are problematic:

*

*You continually speak of the function $\arcsin(\sin(x))$, when in fact the function is $\arcsin \circ \sin$. Or you could speak of the function $g$ defined by $g(x)=\arcsin(\sin(x))$ for all $x$. Notice here how $x$ is a dummy variable that merely serves to illustrate what happens when you plug a number into the function. We could just as well speak of the function $g$ defined by $g(y)=\arcsin(\sin(y))$ for all $y$. It makes no difference. Saying that $\arcsin(\sin(x))$ is a function is an abuse of notation. If you already know and understand this, then feel free to move on to the next bullet point.


*A function is only well-defined if you state what the domain is from the outset. So when you ask, 'what is the domain of $\arcsin(\sin(x))$', you should really be asking 'for what values of $x$ does $\arcsin(\sin(x))$ make sense?'
Now that we have got those technicalities out of the way, we can address the crux of your question. Consider how $\arcsin$ is a function that accepts inputs between $-1$ and $1$. The image of $\sin$ is $[-1,1]$, meaning that the outputs of the sine function always work as inputs to the arcsine function. And since the domain of $\sin$ is $\mathbb{R}$, the domain of $\arcsin \circ \sin$ is also $\mathbb{R}$.
As you have correctly pointed out, $\arcsin$ is differentiable on $(-1,1)$. At the points $(-1,-\pi/2)$ and $(1,\pi/2)$, the graph has a vertical tangent, and so the derivative doesn't exist. This means that $g'(x)$ only makes sense when $\sin x \in (-1,1)$. The $x$-values that we want to exclude are $-\pi/2$ and $\pi/2$, but, taking the periodicity of sine into account, we should exclude $\pi/2 \pm k\pi$ for any integer $k$. Hence, $\arcsin \circ \sin$ is differentiable on $\mathbb{R} \setminus \{x \mid x=\pi/2+k\pi, k \in \mathbb{Z}\}$.
A: Domain is $\mathbb{R}$. It's differentiable on $\mathbb{R} \setminus \{\frac{\pi}{2} + n\pi | n \in \mathbb{Z}\}$. Check this by observing that
$$x \in [-\frac{\pi}{2}, \frac{\pi}{2}] \Rightarrow \arcsin (\sin x) = x  $$
Try to write the formula when $x$ is not in that interval.
