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?

Try to simplify the denominator of f(x): 
Note that by the identity $\cos(2t)=2\cos^2t-1$
$$
\cos(3t)+6\cos(2t)+15\cos t+10=4\cos^3t+12\cos^2t+12\cos t+4=4(\cos t+1)^3
$$
so 
$$
\cos(6x)+6\cos(4x)+15\cos(2x)+10=4(\cos(2x)+1)^3
$$
Therefore
$$
f(x)=\frac{1}{4}\cdot\overbrace{\frac{\cos x}{(\cos(2x)+1)^3}}^{(1)}+\frac{5}{4}\cdot\overbrace{\frac{\cos(3x)}{(\cos(2x)+1)^3}}^{(2)}+\frac{1}{4}\cdot\overbrace{\frac{\cos(5x)}{(\cos(2x)+1)^3}}^{(3)}
$$
Now, in order to compute $f'(x)$ and $f''(x)$ compute $(i)'$ and $(i)''$ for each $i\in\{1,2,3\}$. Then assign $x=0$.
