Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, 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 f₁(x) = cot x.

Accordingly, f₁(x + h) = cot (x + h) 

By first principle,

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

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

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

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

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

= \( \dfrac{-1}{sin^2x}\)

cosec²x=− (cotx)'= - cosxec²x……(2) 

Now, let f₂(x) = cosec x. 

Accordingly, f₂(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(-cosec²x) +cot x(-cosec x cot x) 

= -cosec³ x-cot² x cosec x