Question: What is the coefficient of $x^7$ in the Taylor series expansion, around $x=0$ of the function $f(x)=\sin^{-1}(x)$?
I know how to solve using the Taylor expansion, but in that case I will have to take derivative so many times and the calculation becomes cumbersome. I want to see the reasoning behind the solution given below:
Solution: $\;\;$ $$\frac{f^7(0)}{7!}= \;\text{coefficient of }x^7\;\;\text{ in power series expansion of } \sin^{-1}(x)$$ $$=\frac{1}{7}\cdot(\;\text{coefficient of }x^6\;\text{in power series expansion of }(1-x^2)^{-\frac{1}{2}})\;\;\cdots(*)$$ $$=\frac{1}{7}\cdot\frac{\frac{1}{2}(\frac{1}{2}+1)(\frac{1}{2}+2)}{3!}$$
Can someone please explain me how the step labelled $(*)$ follows? Thank you for the help.
HINT: $$\frac{d \mathrm{sin}^{-1}(x)}{dx} = \frac{1}{\sqrt{1-x^2}}$$
$$\frac{dx^7}{dx} = 7x^6$$