Is Quotient Rule Avoidable?

Relevant wiki: Sum to Product Trigonometric Identities

Simplify numerator as follows:

$\underbrace{(\cos 6x+\cos 4x)}+5\underbrace{(\cos 4x+\cos 2x)}+10\underbrace{(\cos 2x+1)}$

Use $\color{teal}{\cos 2A+\cos 2B=2\cos (A+B)\cos (A-B)\\\text{and }\cos 2x+1=2\cos ^2 x}$ so that numerator is:

$2\color{#3D99F6}{\cos x}\cos 5x+5(2\color{#3D99F6}{\cos x}\cos3x)+10(2\color{#3D99F6}{\cos x}\cdot\cos x)$

We see this is exactly $2\color{#3D99F6}{\cos x}$ times the denominator. Hence we get:

$\large y= 2\color{#3D99F6}{\cos x}$

$\large \therefore \dfrac{\mathrm{d}y}{dx}=\boxed{-2\sin x}$

$\begin{aligned} y & = \frac{\cos 6x + 6\cos 4x + 15\cos 2x + 10}{\cos 5x + 5\cos 3x + 10\cos x} \\ & = \frac{\cos 6x + 5\cos 4x + 10\cos 2x + \cos 4x + 5\cos 2x + 10}{\cos 5x + 5\cos 3x + 10\cos x} \\ & = \frac{\cos 2x(\cos 4x + 5\cos 2x + 10) - \sin 2x(\sin 4x + 5\sin 2x) + \cos 4x + 5\cos 2x + 10}{\cos x(\cos 4x + 5\cos 2x + 10) - \sin x(\sin 4x + 5\sin 2x)} \\ & = \frac{(2\cos^2 x-1)(\cos 4x + 5\cos 2x + 10) - 2\sin x \cos x(\sin 4x + 5\sin 2x) + \cos 4x + 5\cos 2x + 10}{\cos x(\cos 4x + 5\cos 2x + 10) - \sin x(\sin 4x + 5\sin 2x)} \\ & = \frac {2\cos x \left[\cancel{\color{#3D99F6}{\cos x(\cos 4x + 5\cos 2x + 10) - \sin x(\sin 4x + 5\sin 2x)}}\right]}{\cancel{\color{#3D99F6}{\cos x(\cos 4x + 5\cos 2x + 10) - \sin x(\sin 4x + 5\sin 2x)}}} \\ \implies y & = 2 \cos x \\ \implies \frac {dy}{dx} & = \boxed{-2 \sin x} \end{aligned}$