Consider y = cos-1(sin x):
I got dy/dx = +/- 1, but my textbook only mentions -1 as a solution. Here's how they arrived at it:
y = cos-1(sin x) = cos-1(cos(π/2-x)) = π/2 - x
=> dy/dx = -1
I got +/- 1 because I used the chain rule and ended up with a square root factor.
Here's How arrived at the solution:
Plotting the graph of dy/dx confirms that both +1 and -1 are solutions (at mutually exclusive intervals) Graph.
I have seen more similar problems, all involving finding the derivative of some inverse trigonometric function (another example).
I couldn't find a way to find when does the derivative appears as positive or negative. Is there any way to find it (without plotting a graph)?
You have a good solution method right up to the point where you lose track of which sign to write in front of the answer. The solution is simple: do not use $\pm\cos x$ as a substitution for $\sqrt{\cos^2 x}$.
A more accurate substitution is $\sqrt{\cos^2 x} = \lvert \cos x \rvert.$
The answer, in fact, is
$$ \frac{\mathrm dy}{\mathrm dx} = - \frac{\cos}{\sqrt{\cos^2 x}} = - \frac{\cos}{\lvert \cos x \rvert} = -\operatorname{sgn}(\cos x)$$
(where $\operatorname{sgn}$ is the "sign" function), restricted to the domain $\mathbb R \setminus \{ x \mid \cos x = 0\}$. That is, the derivative is defined for all real numbers except the odd multiples of $\frac\pi2,$ where $\cos x = 0$ and the function $\arccos(\sin x)$ is not differentiable.
Note that $\operatorname{sgn}(t)$ is usually defined to return $+1$, $-1$, or $0$ depending on whether $t$ is positive, negative, or zero (respectively), but in this exercise we will only ever see non-zero values of $\cos x$. You could also just leave the answer in the form $-\frac{\cos}{\lvert \cos x \rvert}$, which in my opinion is good enough (perhaps even better, since it makes the domain restriction obvious).