I'm having trouble expressing arccos(x) as a Maclaurin series. I realized that once you differentiate it, the function will just expand rapidly, so is there a easy way approaching this problem.
Yes.
Have you considered the Binomial series - Wikipedia, the free encyclopedia?
Yes
$\displaystyle f(x)=\cos^{-1}(x)$ then
$\displaystyle f'(x)=-\frac{1}{\sqrt{1-x^2}}$
expand this in a power series using the binomial series then integrate
$\displaystyle \binom{k}{n}=\frac{k(k-1)(k-2) \cdot \cdot \cdot (k-n+1)}{n!}$
$\displaystyle (1-x^2)^{-1/2}=1-\frac{1}{2}x^2+\frac{\frac{-1}{2}\frac{-3}{2}}{2!}x^4- \cdot \cdot \cdot =$
$\displaystyle -f'(x)=\sum_{n=0}^{\infty} \binom{k}{n}(-1)^nx^{2n} \iff f'(x)=\sum_{n=0}^{\infty} \binom{k}{n}(-1)^{n+1}x^{2n}$
Integrating both sides we get
$\displaystyle \cos^{-1}(x)=\int f'(x)dx=\int \sum_{n=0}^{\infty} \binom{k}{n}(-1)^{n+1}x^{2n} $
Finally
$\displaystyle f(x)=\sum_{n=0}^{\infty} \binom{k}{n} \frac{(-1)^{n+1}x^{2n+1}}{2n+1}$
This is newtons extention of the binomial theorem to fractional and negative values
$\displaystyle
\binom{k}{n}=\frac{k(k-1)(k-2) \cdot \cdot \cdot (k-n+1)}{n!}$
lets expand a few
$\displaystyle \binom{-\frac{1}{2}}{0} =1$ by definition
$\displaystyle \binom{-\frac{1}{2}}{1} =\frac{\frac{-1}{2}}{1!}$
$\displaystyle \binom{-\frac{1}{2}}{2} =\frac{\frac{-1}{2}\frac{-3}{2}}{2!}$
$\displaystyle \binom{-\frac{1}{2}}{3} =\frac{\frac{-1}{2}\frac{-3}{2}\frac{-5}{2}}{3!}$
and so on