to find coefficient of $x^{7}$ in Taylor series expansion of arcsin(x) Question is to find coefficient of $x^{7}$ in Taylor series expansion of arcsin(x) around x=0. It will be very lengthy to go on computing seven derivatives. Am I missing some trick here. What I did is I compute first derivative which is $ 1/\sqrt{1-x^2}$. Then second derivative is $2x/(1-x^{2})$. While going for third derivative I applied product rule and consider only first term which is $2/(1-x^{2})$ as we are interested to find derivative at 0. Doing this way my only required term at seventh derivative came as $4.2.2/(1-x^{2})^{3}$. But my final answer is not matching with answer key. Any help. Thanks.

$\arcsin''(x)=\frac{x}{(1-x^2)^{3/2}}$ and not what you've written.
The Taylor series of $\frac1{\sqrt{1-x^2}}$ is easy to compute because it is $$(1-x^2)^{-1/2}=\sum_{k=0}^\infty \binom{-1/2}{k} (-x^2)^k=\sum_{k=0}^\infty(-1)^k\binom{-1/2}k x^{2k}$$
Therefore $\arcsin x=\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}\binom{-1/2}k x^{2k+1}$.
