How to find the derivative of every point in the interval? For example:
Find the following derivatives for the given values of $x$.
$$\frac{d}{dx}\arccos (x^2),   x \in (−1, 1).$$
Now finding the derivative is easy: $$=-\frac{2x}{\sqrt{1-x^4}}.$$
Am I suppose to find the derivatives at every point in that interval? If so, do I graph it? Or do I just find the derivative of the function and that's it? I am confused why the $x \in (−1, 1)$, is it because it is undefined?
 A: Why $x\in (-1,1)$ only? It is because $\cos t$ has values between $-1$ and $+1$ only when $t$ is real so $x$ such that $|x|\gt 1$ is not considered. Now the problem  is why $1$ and $-1$ are not considered (Refer Note).  
Also note that it is not always so direct to find derivative of inverse trig. functions so be careful while differentiating these and make sure that you consider only that domain where inverse exists (that is, trig. functions are bijective). 
In your case, you may proceed as below also:
$\frac{d}{dx}\arccos (x^2)$ where $x\in (−1, 1)$ 
$x^2\in [0,1)$
Let $\theta=\arccos (x^2)$, clearly $\theta \in (0,\pi/2]$ (considering principal domain of arccos as $[0,\pi]$) so $\cos\theta =x^2$ 
$\frac{d}{dx}\cos\theta=2x\implies -\sin\theta \frac{d}{dx}(\theta)=2x\implies \frac{d}{dx}(\theta)=-\frac{2x}{\sqrt{1-\cos^2\theta}}=-\frac{2x}{\sqrt{1-x^4}}\tag 1$ 
Note: Let $f(x)=\arccos x^2$,where $x\in (-1,1)$ then by definition , derivative of $f$ at $x=1$$\begin{align}&=\lim_{h\to 0^+}\frac{\arccos (1)-\arccos(1-h)^2}{h}\\&=\lim_{h\to 0^+}\frac{0-\arccos(1-h)^2}{h}\\&=\lim_{h\to 0^+}-\frac{2(1-h)}{\sqrt{1-(1-h)^4}}\\&=\lim_{h\to 0^+}-\frac{2(1-h)}{\sqrt{(1-(1-h)^2)(1+(1-h)^4)}}\to -\infty\end {align}$.
In 3rd line above, L'Hopital's rule has been applied using result obtained in $(1)$.
Similarly can you show that $f'(-1)$ is also not finite?
