Find the derivative of the following function (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\frac{a+b \sin x}{c+d \cos x}$

Let $f(x) = \frac{a+b \sin x}{c+d \cos x}$.
Using the quotient rule,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$:
$f'(x) = \frac{(c+d \cos x) \frac{d}{dx}(a+b \sin x) - (a+b \sin x) \frac{d}{dx}(c+d \cos x)}{(c+d \cos x)^2}$
$f'(x) = \frac{(c+d \cos x)(b \cos x) - (a+b \sin x)(-d \sin x)}{(c+d \cos x)^2}$
$f'(x) = \frac{bc \cos x + bd \cos^2 x + ad \sin x + bd \sin^2 x}{(c+d \cos x)^2}$
Using the identity $\sin^2 x + \cos^2 x = 1$:
$f'(x) = \frac{bc \cos x + ad \sin x + bd(\cos^2 x + \sin^2 x)}{(c+d \cos x)^2}$
$f'(x) = \frac{bc \cos x + ad \sin x + bd}{(c+d \cos x)^2}$