calculus - calculating a higher order derivative

My task is to find the values $f^{(2017)}(0)$ and $f^{(2018)}(0)$ for $f(x)=\frac{arccos(x)}{\sqrt{1-x^2}}$.

Basically, it's about finding the $n^{th}$ derivative of $f$.
So I noticed if I let $g(x)=arccos(x)$, then $f(x)=-g(x)\cdot g'(x)$. I was able to prove by induction that for all $n\geq 2$ and $k\in \Bbb{N}$ the $n^{th}$ derivative of $g'$ is

$$[(g')^{k}]^{(n)}(0)=k\cdot(n-1)\cdot[(g')^{k+2}]^{(n-2)}(0)$$

But even with applying the Leibniz rule to $f=g\cdot g'$ I don't understand how to get the final result. Did I make the wrong approach to the problem or is the general formula above useful? If so, how should I apply it?