How to get nth derivative of $\arcsin x$

Question

I want to calculate the nth derivative of $\arcsin x$. I know $$ \frac{d}{dx}\arcsin x=\frac1{\sqrt{1-x^2}} $$ And $$ \frac{d^n}{dx^n} \frac1{\sqrt{1-x^2}} = \frac{d}{dx} \left(P_{n-1}(x) \frac1{\sqrt{1-x^2}}\right) = \left(-\frac{x}{(1-x^2)^{}} P_{n-1}(x) + \frac{dP_{n-1}}{dx}\right)\frac1{\sqrt{1-x^2}} = P_n(x) \frac1{\sqrt{1-x^2}} $$ Hence we have the recursive relation of $P_n$: $$ P_{n}(x)=-\frac{x}{(1-x^2)^{}} P_{n-1}(x) + \frac{dP_{n-1}}{dx}, \:P_0(x) = 1 $$ My question is how to solve the recursive relation involving function and derivative. I think it should use the generating function, but not sure what it is.

Answer 1

Let $~P_n(x)~=~\dfrac{2^n}{n!}~\Big(\sqrt{1-x^2}\Big)^{2n+1}~\bigg(\dfrac1{\sqrt{1-x^2}}\bigg)^{(n)}.~$ Then its coefficients form the sequence described here.

Answer 2

Outline of a proof that does not use the recursive relation but does yield the correct answer.

Use Faà di Bruno's formula for $n$-fold derivatives in it's "Bell polynomial" form. This omits the question how the tuples affect the final form of the derivatives, so start with:

\begin{equation} \frac{d^n}{dx^n}f(g(x)) = \sum_{k=1}^n f^{(n)}(g(x)) \cdot B_{n,k} (g'(x),g^{(2)}(x),\cdots) \end{equation} with \begin{equation} B_{n,k}(x_1,x_2,\cdots) = \sum_{\sum_i j_i = k, \sum_\ell \ell j_\ell = n} \frac{n!}{\prod_i j_i!} \prod_i\left(\frac{x_i}{i!}\right)^{j_i} \end{equation}

Once this is done, it's useful to think about how the tuples in the Bell's polynomial summation are constrained (hint: only 1 term survives).

The final answer you're looking for is:

\begin{equation} \frac{d^n (1/\sqrt{1-x^2})}{d x^n} = \sum_{k=1}^n (-)^k \left(\frac12 - k\right)_k \left(1-x^2\right)^{-1/2-k} \frac{(2k-n+1)_{2(n-k)} (2x)^{2k-n}}{(n-k)!} \end{equation} where the subbed expressions are Pochhammer symbols.

I think that relating this answer to the expression derived when say $n=4$ is taken or $n=7$ or whatever, you will find that the combinatorial expressions from the explicit $n$ add up in non-intuitive ways to the expression found here, and that in that sense, the setup with the recursive approach should be reconsidered.

Hope someone will find this useful.