If $f(x) = \frac{\cos x + 5\cos 3x + \cos 5x}{\cos 6x + 6\cos4x + 15\cos2x + 10}$then.. If $f(x) = \frac{\cos x + 5\cos 3x + \cos 5x}{\cos 6x + 6\cos4x + 15\cos2x + 10}$ then find the value of $f(0) + f'(0) + f''(0)$.
 I tried differentiating the given. But it is getting too long and complicated. So there must be a way to simplify $f(x)$. What is it?

We can try to simplify the fraction using trigonometric formulae :
$$\cos (x) + 5\cos (3x) + \cos (5x)=\cos(x) (2 \cos(2 x)-1) (2 \cos(2 x)+5)$$ 
and $$\cos (6x) + 6\cos(4x) + 15\cos(2x) + 10=32 \cos^6(x)$$
So $$f(x)=\frac{(2 \cos(2 x)-1) (2 \cos(2 x)+5)}{32\cos^5(x)}=\frac{4\cos^2(2x)+8\cos(2x)-5}{32\cos^5(x)}$$
$$f'(x)=\frac{1}{32}(5 (12 \cos^2(2 x)-5) \tan(x) \sec^5(x)-48 \sin(2 x) \cos(2 x) \sec^5(x))\\=\frac{1}{32}\sec^5(x)(5 (12 \cos^2(2 x)-5) \tan(x)-48 \sin(2 x) \cos(2 x) )$$
But the calculation of $f''(x)$ is still pretty tedious...
