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 ) $cosec \: x \cot x$

Let $f(x) = cosec \: x \cot x$
By Leibnitz product rule,
$f'(x) = cosec\: x( \cot x)'+ (\cot x) cosec \: x'$---------(1)
Let $f_1(x) = \cot x$. Accordingly, $f_1(x+h)=\cot(x+h)$
By first principle,
$f'_1(x) = \lim\limits_{h \to 0} \Large\frac{f_1(x+h)-f_1(x)}{h}$
$= \lim\limits_{h \to 0} \Large\frac{\cot(x+h)-\cot x}{h}$
$= \lim\limits_{ h \to 0} \large\frac{1}{h}$$\bigg[ \Large\frac{\cos (x+h)}{\sin (x+h)}$$ - \Large\frac{\cos x}{\sin x} \bigg]$
$= \lim\limits_{ h \to 0} \large\frac{1}{h}$$\bigg[ \Large\frac{\sin x \cos (x+h)- \cos x \sin (x+h)}{\sin x \sin (x+h)} \bigg] = \lim\limits_{ h \to 0} \large\frac{1}{h}$$ \bigg[ \Large\frac{\sin (x-x-h)}{\sin x \sin (x+h)}\bigg]$
$=\large\frac{1}{\sin x}$$\lim\limits_{ h \to 0} \large\frac{1}{h}$$ \bigg[ \Large\frac{\sin (-h)}{\sin (x+h)} \bigg]$
$=\large\frac{-1}{\sin x}$$\bigg(\lim\limits_{ h \to 0} \large\frac{\sin h}{h}\bigg)$$\bigg(\lim\limits_{ h \to 0} \large\frac{1}{\sin (x+h)}\bigg)$