Derivative of $\sin^{-1}(x)$ I can find this using the fact that $\sin(\sin^{-1}(x)) = x$, for all $x\in[-1,1].$
Now, differentiate. 
$$\frac{d}{d\sin^{-1}(x)}\sin(\sin^{-1}(x))\cdot \frac{d}{dx} \sin^{-1}(x)= \frac{d}{dx} x= 1$$
$$\cos(\sin^{-1}(x))\cdot \frac{d}{dx} \sin^{-1}(x) = 1$$
$$\frac{d}{dx} \sin^{-1}(x) = \frac{1}{\cos(\sin^{-1}(x))}$$
$$\frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1-\sin^2(\sin^{-1}(x))}}$$
$$\frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}}$$
However, what if I wanted to differentiate this like $\ \sin^{-1}(\sin(x))$ without knowing the fact that $\ \frac{d}{dx}\sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}}$ ? Is there a solution for it? I keep getting stuck at a certain step when I try this...

You could do it like this:
first, define $f$ by $f(x) = \sin^{-1}x,$
where $-1 \leq x \leq 1$ and 
$-\frac\pi2 \leq \sin^{-1}x \leq \frac\pi2.$
(Recall that $\sin^{-1}$ is not a real function for $|x| > 1$
and that $\sin$ is not a one-to-one function, so we have to choose
an appropriate domain and range of $\sin^{-1}$; these are the 
usual choices.)
Then for $-\frac\pi2 \leq \theta \leq \frac\pi2,$
$$
f(\sin \theta) = \theta. \tag1
$$
Take the derivative of each side of $(1)$ with respect to $\theta$:
$$
\frac{d}{d\theta}\left(f(\sin\theta)\right) = 1. \tag2
$$
Evaluate the left-hand side of $(2)$ using the chain rule:
$$
\frac{d}{d\theta}\left(f(\sin\theta)\right)
 = f'(\sin\theta)\frac{d}{d\theta}\left(\sin\theta\right)
 = f'(\sin\theta) \cos\theta. \tag3
$$
Combine $(2)$ and $(3)$:
$$
f'(\sin\theta) \cos\theta = 1. \tag4
$$
Divide by $\cos\theta$ on both sides of $(4)$:
$$
f'(\sin\theta) = \frac{1}{\cos\theta}. \tag5
$$
Let $x = \sin\theta.$ Then $1 - x^2 = \cos^2\theta,$
and since $-\frac\pi2 \leq \theta \leq \frac\pi2,$
it follows that $\cos\theta\geq 0,$ so we have
$\cos\theta = \sqrt{1 - x^2}.$
Making these substitutions in $(5)$,
$$
f'(x) = \frac{1}{\sqrt{1 - x^2}}. 
$$
But $f'(x)=\frac{d}{dx}\sin^{-1}x,$ so
$$
\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}.
$$
