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 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
