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: we can simplify the fraction as $$\frac{2\cos3x\cos2x+5\cos3x}{2\cos^23x-1+6[2\cos3x\cos x]+9\cos2x+10}$$
$$=\frac{(2\cos2x+5)\cos3x}{2\cos^23x+12\cos3x\cos x+18\cos^2x}$$
$$=\frac{(2\cos2x+5)\cos3x}{2(\cos3x+3\cos x)^2}$$
$$=\frac{[2(2c^2-1)+5](4c^3-3c)}{2(4c^3)^2}$$
$$=\frac{(4c^2+3)(4c^2-3)}{32c^5}$$
$$=\frac 12\sec x-\frac{9}{32}\sec^5x$$
Now you can differentiate this twice quite easily
A: $f(x) = \frac{u(x)}{v(x)}$
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.
A: 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 $$A_0=\sum _{i=0}^m a_i\qquad A_1=-\frac{1}{2} \sum _{i=0}^m i^2 a_i\qquad B_0=\sum _{i=0}^n b_i\qquad B_1=-\frac{1}{2} \sum _{i=0}^n i^2 b_i\qquad$$
A: There doesn't appear to be a way to simplify f(x) to make it easy to differentiate, however one way to make the differentiation less labour intensive is to express $f'(x)$ and $f''(x)$ in terms of the derivatives of the numerator and denominator.
Beginning from  $f(x)=\frac{u(x)}{v(x)}$
Using the quotient rule
$f'(x)=\frac{vu'-uv'}{v^2}$
Applying the quotient, product, and chain rules
$f''(x)=\frac{v^2(v'u'+vu''-v'u'-uv'')-2vv'(vu'-uv')}{v^4}$
If you calculate $u(0), u'(0),u''(0)$ and $v(0), v'(0),v''(0)$ then it is straightforward to find the second derivative of $f$
A: 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...
A: 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 $x=0$.
