Power Series for $\arcsin(x)$ and $\arcsin(x^2)$

Question

I am having some trouble solving this question:

Determine a power series entered at 0 for: $$f(x) = \arcsin(x^2)$$ Hence, write an expression for the $kth$ derivative of $\arcsin(x)$, where $k \in \mathbb{N} $.

Side note: I am able to derive the expression for the Maclaurin Series of $\arcsin(x)$, but I don't really understand how to use that information in obtaining the $kth$ derivative of $\arcsin(x)$.

Any help is appreciated! Thanks a lot.

https://math.stackexchange.com/questions/1550934/how-to-get-nth-derivative-of-arcsin-x — zkutch, Jun 20 '20 at 06:54
I don't really think either of these posts helps me. I am able to get the Maclaurin Expansion for $\arcsin(x^2)$, but I can't seem to relate it to the derivatives of $\arcsin(x)$. — GanTheMan, Jun 20 '20 at 06:59
Take or build the expansion of $\arcsin(y)$ and make $y=x^2$. — Claude Leibovici, Jun 20 '20 at 07:50
Ok, I think I understand what I have to do now. Thanks to all of you! — GanTheMan, Jun 20 '20 at 09:40

score 0 · Answer 1 · answered Jun 20 '20 at 13:27

$$\arcsin(x) = x + \sum\limits_{k = 1}^\infty {\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}}\frac{{{x^{2k + 1}}}}{{2k + 1}}}$$ $$f(x)=\arcsin(x^2)=x + \sum\limits_{k = 1}^\infty {\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}}\frac{{{x^{4k + 2}}}}{{2k + 1}}}$$ for ease im going to say $s=\frac{d}{dx}$ and so: $$sf(x)=1+\sum\limits_{k = 1}^\infty {\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}}\frac{{{x^{4k + 1}}}}{{2k + 1}}}(4k+2)=1+2\sum\limits_{k = 1}^\infty {\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}}{{x^{4k + 1}}}}$$ now: $$s^2f(x)=2\sum\limits_{k = 1}^\infty (4k+1){\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}}{{x^{4k}}}}$$ now any higher derivates you can define in terms of a factorial

Power Series for $\arcsin(x)$ and $\arcsin(x^2)$

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