Find the derivative of the following function: $\frac{4x + 5 \sin x}{3x + 7 \cos x}$

Let $f(x) = \frac{4x + 5 \sin x}{3x + 7 \cos x}$.
Using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$:
$f'(x) = \frac{(3x + 7 \cos x) \frac{d}{dx}(4x + 5 \sin x) - (4x + 5 \sin x) \frac{d}{dx}(3x + 7 \cos x)}{(3x + 7 \cos x)^2}$
$f'(x) = \frac{(3x + 7 \cos x)(4 + 5 \cos x) - (4x + 5 \sin x)(3 - 7 \sin x)}{(3x + 7 \cos x)^2}$
Expanding the numerator:
$= \frac{(12x + 15x \cos x + 28 \cos x + 35 \cos^2 x) - (12x - 28x \sin x + 15 \sin x - 35 \sin^2 x)}{(3x + 7 \cos x)^2}$
$= \frac{12x + 15x \cos x + 28 \cos x + 35 \cos^2 x - 12x + 28x \sin x - 15 \sin x + 35 \sin^2 x}{(3x + 7 \cos x)^2}$
Since $\cos^2 x + \sin^2 x = 1$,we have $35(\cos^2 x + \sin^2 x) = 35$:
$= \frac{35 + 15x \cos x + 28 \cos x + 28x \sin x - 15 \sin x}{(3x + 7 \cos x)^2}$