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.
$$f'(x) =  \frac {1'(1-2\cos x)-1(1-2\cos^2x)'}{(1-2\cos^2x)^2} = 
\frac {1-2\cos^2x-(1-(2\cos^2x)')}{(1-2\cos^2x)^2} = $$
$$=\frac {-2\cos^2x + 2(2\cos x(\cos x)')}{(1-2\cos^2x)^2} = 
\frac {-2\cos^2x+2(-2\sin x\cos x)}{(1-2\cos^2x)^2} = $$
$$\frac {-2\cos^2x-4\sin x\cos x}{(1-2\cos^2x)^2}$$

$f(x) = \frac{1}{1-2cos^2x}$
The result of $f'(x)$ should be equal $f'(x) = \frac{-4cosxsinx}{(1-2cos^2x)^2}$
$$
f'(x) = 
\frac {0-1(1-2cos^2x)'}  {(1-2cos^2x)^2} = 
\frac {-4sinxcosx}          {(1-2cos^2x)^2}
$$
