Question

Solve:$\frac{d}{dx}\left(cosecx\right)=?$

Answer 1

Given,

$\frac{d}{dx}\left(cosecx\right)=\frac{d}{dx}\left(\frac{1}{\sin x}\right)$

We know the quotient rule of differentiation,

$\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$

Now, let $u=1,\sin x=v$

$\therefore \frac{d}{dx}\left(\frac{1}{\sin x}\right)=\frac{\sin x\frac{d}{dx}\left(1\right)-1\frac{d}{dx}\left(\sin x\right)}{{\left(\sin x\right)}^2}=\frac{\sin x\left(0\right)-\cos x}{{\left(\sin x\right)}^2}=\frac{-\cos x}{{\left(\sin x\right)}^2}=\frac{-\cos x}{\sin x}\cdot \frac{1}{\sin x}=-\cot xcosecx\therefore \frac{d}{dx}\left(cosecx\right)=-\cot xcosecx.$