How to calculate derivative of $f(x) = \frac{1}{1-2\cos^2x}$? $$f(x) = \frac{1}{1-2\cos^2x}$$
The result of $f'(x)$ should be equals
$$f'(x) = \frac{-4\cos x\sin x}{(1-2\cos^2x)^2}$$
I'm trying to do it in this way but my result is wrong.
$$f'(x) =  \frac {1'(1-2\cos x)-1(1-2\cos^2x)'}{(1-2\cos^2x)^2} = 
\frac {1-2\cos^2x-(1-(2\cos^2x)')}{(1-2\cos^2x)^2} = $$
$$=\frac {-2\cos^2x + 2(2\cos x(\cos x)')}{(1-2\cos^2x)^2} = 
\frac {-2\cos^2x+2(-2\sin x\cos x)}{(1-2\cos^2x)^2} = $$
$$\frac {-2\cos^2x-4\sin x\cos x}{(1-2\cos^2x)^2}$$

The problem is in this step, from here
$$f'(x) =  \frac {1'(1-2\cos x)-1(1-2\cos^2x)'}{(1-2\cos^2x)^2} $$
to here
$$\frac {1-2\cos^2x-(1-(2\cos^2x)')}{(1-2\cos^2x)^2}$$
because $$1'=0.$$
