Differentiate the function $y=\cos^{-1}(\sin x)$ w.r.t $x$ 
Find $y'$ if $y=\cos^{-1}(\sin x)$

It is solved as
$$
y=\cos^{-1}(\cos(\pi/2-x))=\frac{\pi}{2}-x\\
y'=-1
$$
My Attempt
The function is defined for all real numbers.
$$
y'=\frac{-\cos x}{\sqrt{1-\sin^2x}}=\frac{-\cos x}{\sqrt{\cos^2x}}=\frac{-\cos x}{|\cos x|}=\pm1
$$
Even in the first method isn't it the right way to solve 
$$
y=\cos^{-1}(\cos(\pi/2-x))=\frac{\pi}{2}-x\\\implies\cos y=\cos(\pi/2-x)\\
\implies y=2n\pi\pm\big(\frac{\pi}{2}-x\big)\\
\implies y'=\pm1
$$
Is $y'=-1$ a complete solution ?
 A: The function $\cos\mid_{[0,\pi]}:[0,\pi]\to[-1,1]$ is bijective and has the inverse $\arccos:[-1,1]\to [0,\pi]$ and $\arccos(\cos(x))=x$ for all $x\in[0,\pi]$ while $\cos(\arccos(y))=y$ for all $y\in[-1,1]$. Further $\cos\mid_{[0,\pi]}'(0)=\cos\mid_{[0,\pi]}'(\pi)=0$ and therefore $\arccos$ is just differentiable on $(-1,1)$.
Hence $\arccos(\cos(\frac{\pi}2-x))=\frac{\pi}2-x$ holds just if $\frac{\pi}2-x\in [0,\pi]$, so $x\in \left[-\frac{\pi}2,\frac{\pi}2\right]$.
That's why you should use the chain rule and note that $y$ is not differentiable if $\sin(x)=\pm1$ which is when $x=\frac{\pi}2+k\pi$ for some $k\in\mathbb Z$. Then, your computation are correct:
$$
y'=-\frac{\cos(x)}{|\cos(x)|}.
$$
But you can't just say $y'=\pm 1$. There is just one derivative and it depends on $x$.
$$
y'=-\frac{\cos(x)}{|\cos(x)|}=\begin{cases}
-1 & x\in \left(-\frac{\pi}2+2k\pi,\frac{\pi}2+2k\pi\right)\\
1 & x\in \left(\frac{\pi}2+2k\pi,\frac{3\pi}2+2k\pi\right)
\end{cases}
$$
where $k\in\mathbb Z$.

If you plot the graph, then the result isn't even suprising:
https://www.desmos.com/calculator/odco70auib
A: Hint:
$\cos y=\sin x,$ and $0\le y\le\pi\ \ \ \ (1)$
$y=2k\pi\pm(\pi/2-x)$ where $k$ is an integer satisfying $(1)$
What if $0\le 2k\pi+\pi/2-x\le\pi,?\le x\le?$
