Differentiate each of the following functions from the first principal :

We have to find the derivative of $\log \operatorname{cosec} x$ with the first principle method, so,

$f(x)=\log \operatorname{cosec} x$

by using the first principle formula, we get,

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

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\log \operatorname{cosec}(x+h)-\log \operatorname{cosec} x}{h}$

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\log \sin x-\log \sin (x+h)}{h}$

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\log \frac{\sin x}{\sin (x+h)}}{h}$

[By using $\log a-\log b=\log \frac{a}{b}$ ]

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

[adding and subtracting 1 ]

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

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

[Rationalising]

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

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{2 \cos \frac{2 x+h}{2} \sin \frac{-h}{2}}{h \sin (x+h)}$

$\left[\sin C-\sin D=2 \sin \frac{C-D}{2} \cos \frac{C+D}{2}\right]$

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{-2 \cos \frac{2 x+h}{2} \sin \frac{-h}{2}}{(-1) h \sin (x+h)}$

[By using $\left.\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$

$f^{\prime}(x)=-\cot x$