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?

$f(x) = \frac{u(x)}{v(x)}$
so $f(x) v(x) = u(x)$
Differentiating gives: $f'(x) v(x) + v'(x) f(x) = u'(x)$ 
Differentiating again: $f''(x) v(x) + f'(x) v'(x) + v''(x)f(x)+f'(x)v'(x) = u''(x)$ 
Setting $x=0$ and noticing $u'(0)=v'(0)=0$ (only $\sin$ terms appear in the derivatives) gives:
$f'(0)v(0)=0 \Rightarrow f'(0)=0$ 
Then $f''(0)v(0)+v''(0)f(0)=u''(0)$ and finally:  
$f''(0)=\frac{u''(0)-v''(0)f(0)}{v(0)}$
I'll let you compute $f(0)$,$v(0)$, $u''(0)$ and $v''(0)$ which is elementary.
