Differentiate the following functions:

Solution:

Formulae: -

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

$\frac{d}{d x} \cos x=-\sin x$

$\frac{d}{d x} \sec x=\sec x \tan x$

$\frac{d}{d x} \operatorname{cosecx}=-\operatorname{cosec} x \cot x$

$\frac{d}{d x} \tan x=\sec ^{2} x$

$\frac{d}{d x} \sin x=\cos x$

$\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{k}=0, \mathrm{k}$ is constant

(i) $4 \cot x-\frac{1}{2} \cos x+\frac{2}{\cos x}-\frac{3}{\sin x}+\frac{6 \cot x}{\operatorname{cosec} x}+9$

$=4 \cot x-\frac{1}{2} \cos x+2 \sec x-3 \operatorname{cosec} x+6 \cos x+9$

Differentiating with respect to $x$,

$\frac{d}{d x}\left(4 \cot x-\frac{1}{2} \cos x+2 \sec x-3 \operatorname{cosec} x+6 \cos x+9\right)$

$=4\left(-\operatorname{cosec}^{2} x\right)-\frac{1}{2}(-\sin x)+2 \sec x \times \tan x-3(-\operatorname{cosec} x \times \cot x)+6(-\sin x)+$ 0

$=-4 \operatorname{cosec}^{2} x+\frac{1}{2} \sin x+2 \sec x \tan x+3 \operatorname{cosec} x \cot x-6 \sin x$

(ii) $-5 \tan x+4 \tan x \cos x-3 \cot x \sec x+2 \sec x-13$

$=-5 \tan x+4 \sin x-3 \operatorname{cosec} x+2 \sec x-13$

Differentiating with respect to $\mathrm{x}$,

$\frac{d}{d x}(-5 \tan x+4 \sin x-3 \operatorname{cosec} x+2 \sec x-13)$

$=-5 \sec ^{2} x+4 \cos x-3(-\operatorname{cosec} x \cot x)+2 \sec x \tan x-0$

$=-5 \sec ^{2} x+4 \cos x+3 \operatorname{cosec} x \cot x+2 \sec x \tan x$