Question

Use the quotient rule, or otherwise, to prove that
$\frac{d}{dx}\left(\text{cosec}hx\right)=-\text{cosec}hx\cot hx$

Answer 1

We know that $\text{cosech}x=\frac{1}{\sin hx}$
Hence $\frac{d}{dx}\text{cosech}x=\frac{d}{dx}\frac{1}{\sin hx}$
Differentiate the given equation with respect to x using quotient rule
$\frac{d}{dx}\frac{1}{\sin hx}=\frac{\left[\sin hx\right(d/dx\left(1\right))-(1\left)\right(d/dx\left(\sin hx\right)]}{{\sinh }^{2}x}=\frac{0-\cosh x}{{\sinh }^{2}x}=-\frac{-\cosh x}{{\sinh }^{2}x}=-\frac{-\cosh x}{\sinh x\sinh x}=-\frac{\cosh x}{\sinh x}\frac{1}{\sinh x}=-\coth x\text{cosec}hx$
Hence proved that $\frac{d}{dx}\text{cosech}x=-\coth x\text{cosec}hx$