Find the derivative of the following functions

Solution:

Let $f(x)=\operatorname{cosec} x \cot x$

By Leibnitz product rule,

$f^{\prime}(x)=\operatorname{cosec} x(\cot x)^{\prime}+\cot x(\operatorname{cosec} x)^{\prime}$ $\ldots(1)$

Let $f_{1}(x)=\cot x$. Accordingly, $f_{1}(x+h)=\cot (x+h)$

By first principle,

$f_{1}^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f_{1}(x+h)-f_{1}(x)}{h}$

$=\lim _{h \rightarrow 0} \frac{\cot (x+h)-\cot x}{h}$

$=\lim _{h \rightarrow 0} \frac{1}{h}\left(\frac{\cos (x+h)}{\sin (x+h)}-\frac{\cos x}{\sin x}\right)$

$=\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{\sin x \cos (x+h)-\cos x \sin (x+h)}{\sin x \sin (x+h)}\right]$

$=\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{\sin (x-x-h)}{\sin x \sin (x+h)}\right]$

$=\frac{1}{\sin x}, \lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{\sin (-h)}{\sin (x+h)}\right]$

$=\frac{-1}{\sin x} \cdot\left(\lim _{h \rightarrow 0} \frac{\sin h}{h}\right)\left(\lim _{h \rightarrow 0} \frac{1}{\sin (x+h)}\right)$

$=\frac{-1}{\sin x} \cdot 1 \cdot\left(\frac{1}{\sin (x+0)}\right)$

$=\frac{-1}{\sin ^{2} x}$

$=-\operatorname{cosec}^{2} x$ $.(2)$

Now, let $f_{2}(x)=\operatorname{cosec} x$. Accordingly, $f_{2}(x+h)=\operatorname{cosec}(x+h)$

By first principle,

$f_{2}^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f_{2}(x+h)-f_{2}(x)}{h}$'

$=\lim _{h \rightarrow 0} \frac{1}{h}[\operatorname{cosec}(x+h)-\operatorname{cosec} x]$

$=\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{1}{\sin (x+h)}-\frac{1}{\sin x}\right]$

$=\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{\sin x-\sin (x+h)}{\sin x \sin (x+h)}\right]$

$=\frac{1}{\sin x} \cdot \lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{2 \cos \left(\frac{x+x+h}{2}\right) \sin \left(\frac{x-x-h}{2}\right)}{\sin (x+h)}\right]$

$=\frac{1}{\sin x} \cdot \lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{2 \cos \left(\frac{2 x+h}{2}\right) \sin \left(\frac{-h}{2}\right)}{\sin (x+h)}\right]$

$=\frac{-1}{\sin x} \cdot \lim _{h \rightarrow 0} \frac{\sin \left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)} \cdot \lim _{h \rightarrow 0} \frac{\cos \left(\frac{2 x+h}{2}\right)}{\sin (x+h)}$

$=\frac{-1}{\sin x} \cdot 1 \cdot \frac{\cos \left(\frac{2 x+0}{2}\right)}{\sin (x+0)}$

$=\frac{-1}{\sin x} \cdot \frac{\cos x}{\sin x}$

$=-\cos \operatorname{ec} x \cdot \cot x$

$\therefore(\operatorname{cosec} x)^{\prime}=-\operatorname{cosec} x \cdot \cot x$...(3)

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

$f^{\prime}(x)=\operatorname{cosec} x\left(-\operatorname{cosec}^{2} x\right)+\cot x(-\operatorname{cosec} x \cot x)$

$=-\operatorname{cosec}^{3} x-\cot ^{2} x \operatorname{cosec} x$