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?

A "simple" way could be to consider Taylor series; start using $$\cos(y)=1-\frac{y^2}{2}+\frac{y^4}{24}+O\left(y^6\right)$$ and replace $y$ by what is required.
Doing it, the numerator should be $$7-\frac{71 x^2}{2}+\frac{1031 x^4}{24}+O\left(x^6\right)$$ and the denominator $$32-96 x^2+128 x^4+O\left(x^6\right)$$ Now, long division. 
When you get the result, remember that $$f(x)=f(0)+x f'(0)+\frac{1}{2} x^2 f''(0)+O\left(x^3\right)$$ Just identify the terms to get $f(0)$, $f'(0)$ and $f''(0)$.
Edit
As Yves Daoust commented, it will be sufficient to use $$\cos(y)=1-\frac{y^2}{2}+O\left(y^4\right)$$ This makes the work much faster.
Update
We may suppose that, for this problem, the coefficients were selected in order to allow nice trigonometric simplifications such as David Quinn's ones.
Let us take the most general case of $$f(x)=\frac{\sum_{i=0}^m a_i \cos(i x)}{\sum_{i=0}^n b_i \cos(i x)}$$ Using the same approach as in this answer we shall have $$f(x)=\frac{A_0+A_1 x^2+O\left(x^4\right)}{B_0+B_1 x^2+O\left(x^4\right)}=\frac{A_0}{B_0}+\frac{A_1B_0-A_0B_1}{B_0^2}x^2+O\left(x^4\right)$$ using $$A_0=\sum _{i=0}^m a_i\qquad A_1=-\frac{1}{2} \sum _{i=0}^m i^2 a_i\qquad B_0=\sum _{i=0}^n b_i\qquad B_1=-\frac{1}{2} \sum _{i=0}^n i^2 b_i\qquad$$
