Finding the Taylor series of $\arcsin(1-x)$ I'm trying to calculate the Taylor series of $\arcsin(1-x)$ about $x=0$.  I'm having trouble because I can't compute the derivative there.
I can see the correct solution on WolframAlpha (http://www.wolframalpha.com/input/?i=taylor+series+arcsin(1-x)), but I'd like to know how to derive it.  I found a similar question regarding the Taylor series for $\arccos(1-x)$ (Taylor expansion of $\arccos(1-x)$ around $x=0$ to two terms), but the derivation takes advantage of the fact that $\arccos(1-x) << 1$, which is not the case for arcsin.
Any help would be greatly appreciated.
 A: 
Let $f(x)=\arcsin(1-x)$ for $x\in[0,2]$. 
Since the derivative of $f(x)=O\left( x^{-1/2}\right)$ for $x\sim 0$, we let $t=x^{1/2}$ and $g(t)=\arcsin(1-t^2)$.  
We will now develop the first few terms of the Taylor series for $g(t)$ around $t=0$.  


We have for the first derivative $g^{(1)}(t)$
$$\begin{align}
g^{(1)}(t)&=-\frac{2t}{\sqrt{1-(1-t^2)^2}}\\\\
&=-\frac{2}{\sqrt{2-t^2}}\tag 1
\end{align}$$   

Differentiating the right-hand side of $(1)$, we obtain the second derivative, $g^{(2)}(t)$
$$\begin{align}
g^{(2)}(t)&=-\frac{2t}{(2-t^2)^{3/2}}\tag 2
\end{align}$$   

Continuing, we have for $g^{(3)}(t)$
$$\begin{align}
g^{(3)}(t)&=-\frac{4(t^2+1)}{(2-t^2)^{5/2}}\tag 3
\end{align}$$   

And finally, we have for $g^{(4)}(t)$
$$\begin{align}
g^{(4)}(t)&=-\frac{12t(t^2+3)}{(2-t^2)^{7/2}}\tag 4
\end{align}$$   

We evaluate $(1)-(4)$ at $t=0$ and form the expansion
$$\bbox[5px,border:2px solid #C0A000]{\arcsin(1-x)=\frac{\pi}{2}-\sqrt{2}x^{1/2}-\frac{\sqrt{2}}{12}x^{3/2}+O\left(x^{5/2}\right)}$$ 
