Solve this

Solution:

To find: Differentiation of cotx

Formula used: (i) $\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

(ii) $\frac{d \sin x}{d x}=\cos x$

(iii) $\frac{d \cos x}{d x}=-\sin x$

We can write cotx as $\frac{\cos x}{\sin x}$

Let us take u = cosx and v = sinx

$u^{\prime}=(\cos x)^{\prime}=-\sin x$

$v^{\prime}=(\sin x)^{\prime}=\cos x$

Putting the above obtained values in the formula:-

$\left(\frac{\mathrm{u}}{\mathrm{v}}\right)^{\prime}=\frac{\mathrm{u}^{\prime} \mathrm{v}-\mathrm{uv}^{\prime}}{\mathrm{v}^{2}}$ where $\mathrm{v} \neq 0$ (Quotient rule)

$\left(\frac{\cos x}{\sin x}\right)^{\prime}=\frac{(-\sin x)(\sin x)-(\cos x)(\cos x)}{(\sin x)^{2}}$

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

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

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

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

Ans)

$-\operatorname{cosec}^{2} x$

(ii)

To find: Differentiation of secx

Formula used: (i) $\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

(ii) $\frac{d \cos x}{d x}=-\sin x$

We can write secx as $\frac{1}{\cos x}$

Let us take u = 1 and v = cosx

$u^{\prime}=(1)^{\prime}=0$

$v^{\prime}=(\cos x)^{\prime}=-\sin x$

Putting the above obtained values in the formula:

$\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

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

$=\frac{\sin x}{\cos ^{2} x}$

$=\sec x \tan x$

Ans).

$-\operatorname{cosec}^{2} x$