<s>[INST] Maclaurin expansion of $\arccos(1-2x^2)$ Maclaurin expansion of $\arccos(1-2x^2)$
This is what I tried.
$f'(x)=2(1-x^2)^{-1/2} \\
f''(x)=2(1-x^2)^{-3/2}+3 \cdot 2 x^2(1-x^2)^{-5/2} \\
f^{(3)}(x)=18x(1-x^2)^{-5/2}+2\cdot 3\cdot 5x^3(1-x^2)^{-7/2} \\
f^{(4)}(x)=18(1-x^2)^{-5/2}+180x^2(1-x^2)^{-7/2}+2\cdot 3\cdot 5\cdot 7x^4(1-x^2)^{-9/2}$
From this I get $f'(0)=2 $, $f''(0)=0 $, $f^{(3)}(0)=2 $, $f^{(4)}(0)=0 $
But I don't know how to find a general term. Can this be solved in easier steps? 
 [/INST] If you check your differentiation very carefully, you will find that $f(x)=\arccos(1-2x^2)$ is not continuously differentiable at $x=0$, so the MacLaurin expansion does not exist.
Details: if $x\ne0$ we have
$$\frac{d}{dx}\arccos(1-2x^2)=\frac{4x}{\sqrt{1-(1-2x^2)^2}}
  =\frac{2x}{|x|\sqrt{1-x^2}}\ .$$
So
$$\lim_{x\to0^+}f'(x)=2\quad\hbox{but}\quad \lim_{x\to0^-}f'(x)=-2\ .$$
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