Find the derivative of the following functions (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): $\frac{4 x+5 \sin x}{3 x+7 \cos x}$
Let $f(x)=\frac{4 x+5 \sin x}{3 x+7 \cos x}$
By quotient rule,
$f^{\prime}(x)=\frac{(3 x+7 \cos x) \frac{d}{d x}(4 x+5 \sin x)-(4 x+5 \sin x) \frac{d}{d x}(3 x+7 \cos x)}{(3 x+7 \cos x)^{2}}$
$=\frac{(3 x+7 \cos x)\left[4 \frac{d}{d x}(x)+5 \frac{d}{d x}(\sin x)\right]-(4 x+5 \sin x)\left[3 \frac{d}{d x} x+7 \frac{d}{d x} \cos x\right]}{(3 x+7 \cos x)^{2}}$
$=\frac{(3 x+7 \cos x)(4+5 \cos x)-(4 x+5 \sin x)(3-7 \sin x)}{(3 x+7 \cos x)^{2}}$
$=\frac{12 x+15 x \cos x+28 \cos x+35 \cos ^{2} x-12 x+28 x \sin x-15 \sin x+35 \sin ^{2} x}{(3 x+7 \cos x)^{2}}$
$=\frac{15 x \cos x+28 \cos x+28 x \sin x-15 \sin x+35\left(\cos ^{2} x+\sin ^{2} x\right)}{(3 x+7 \cos x)^{2}}$
$=\frac{35+15 x \cos x+28 \cos x+28 x \sin x-15 \sin x}{(3 x+7 \cos x)^{2}}$