Consider the function g(x)=cos(f(x)) and its derivative. How are g′(x) and f′(x) related to each other? Find the derivative of the function:
$$f(x)=\left(\arccos \left(\sqrt{1-x^2}\right)\right)$$

Ok for this I would use the chain rule to get $f'(x)=\frac{x}{\sqrt{x^2}\sqrt{1-x^2}}$
However the question says this:

Hint: Consider the function $g(x) = cos(f(x))$ and its derivative. How are $g' (x)$ and $f' (x)$ related to each other?

Now this part I don't get, what do they mean? Does this mean I can't use chain rule to get the derivative?

 A: $g(x)=\sqrt {1-x^2}$
Now find $g'( x)$ and compare it with your $f'(x)$
A: I assume they mean the implicit differentiation. 
Consider the function $g(x) = cos(f(x))$, where $g(x)=\sqrt{1-x^2}$. Then:
$$\begin{align}g'(x)&=-\sin (f(x))\cdot f'(x) \Rightarrow \\
f'(x)&=-\frac{g'(x)}{\sin (f(x))}=\\
&=-\frac{-\frac{x}{\sqrt{1-x^2}}}{\sin(\arccos{\sqrt{1-x^2}})}=\\
&=-\frac{-\frac{x}{\sqrt{1-x^2}}}{\sqrt{1-\cos^2(\arccos{\sqrt{1-x^2}})}}=\\
&=\frac{x}{\sqrt{x^2}\sqrt{1-x^2}}.\end{align}$$
A: $$f(x)=\arccos\left(\sqrt{1-x^2}\right)$$
$$\sqrt{1-x^2}=\cos\left(f(x)\right)$$
so:
$$g(x)=\sqrt{1-x^2}$$
$$g'(x)=\frac{x}{\sqrt{1-x^2}}$$
so:
$$\frac{x}{\sqrt{1-x^2}}=-f'(x)\sin\left(f(x)\right)$$
A: Ok, it's kind of stupid, but I think this is what your question is "trying" to show.
We have \begin{equation} f(x) = \arccos{\sqrt{1-x^2}}. \end{equation}
Then
\begin{equation}g(x) \dot= \cos(f(x))= \sqrt{1-x^2}. \end{equation}
Taking derivatives,
\begin{equation}g' = -\sin{(f(x))}f'(x) = \frac{-x}{\sqrt{1-x^2}}. \end{equation}
Now, use trigonometry to derive$$\sin{f(x)}= \sin{\arccos{\sqrt{1-x^2}}}=|x|.$$
Divide to get the answer $$f'(x) = \frac{x}{|x|\sqrt{1-x^2}}.$$
I guess the benefit is to show that you can do the problem without knowing explicitly how to differentiate $\arccos{x}$.
