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,

\( f'(x)=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}\)

\( f'(x)=\lim\limits_{h \to 0}\dfrac{cot(x+h)-cot(x)}{h}\)

\(\lim\limits_{h \to 0}\dfrac{1}{h}(\dfrac{cot(x+h)}{sin(x+h)}-\dfrac{cos(x)}{sinx})\)

= \( \dfrac{-1}{sinx} (\lim\limits_{h \to 0} \dfrac{sinh}{h} )[ \lim\limits_{h \to 0}\dfrac{1}{sin(x+h)}] \)

= \( \dfrac{-1}{sinx}.1.( \lim\limits_{h \to 0}\dfrac{1}{sin(x+0)})\)

= \( \dfrac{-1}{sin^2x}\)

cosec^{2}x=− (cotx)'= - cosxec^{2}x……(2)

Now, let f_{2}(x) = cosec x.

Accordingly, f_{2}(x + h) = cosec(x + h)

By first principle,

\( f_2'(x)=\lim\limits_{h \to 0}\dfrac{f_2(x+h)-f_2(x)}{h} \)

\(\lim\limits_{h \to 0}\dfrac{1}{h}(\dfrac{1}{sin(x+h)}-\dfrac{1}{sinx})\)

\( \lim\limits_{h \to 0}\dfrac{1}{h}(\dfrac{sinx-sin(x+h)}{sinxsin(x+h)})\)

= \( \dfrac{-1}{sinx}.1.\dfrac{(\dfrac{2x+h}{2})}{sin(x+0)}\)

= \( \dfrac{-1}{sinx}.\dfrac{cosx}{sinx}\)

(cosec x)’=-cosec x. cot x

From (1), (2), and (3), we obtain

f'(x) = cosec x(-cosec^{2}x) +cot x(-cosec x cot x)

= -cosec^{3 }x-cot^{2} x cosec x