Question

Use the quotient rule, or otherwise, to prove that
$\frac{d}{dx}\left(\cot hx\right)=-co\sec h^{2}x$

Answer 1

$\cot hx=\frac{e^x+e^{-x}}{e^x-e^{-x}}$

now,

$\frac{d\left(\frac{e^x+e^{-x}}{e^x-e^{-x}}\right)}{dx}=\frac{(e^x-e^{-x})(e^x-e^{-x})-(e^x+e^{-x})(e^x+e^{-x})}{(e^x-e^{-x}{)}^2}$

$-\frac{4}{(e^x-e^{-x}{)}^2}=-co\sec h^{2}x$

hence,

$\frac{d\left(\cot hx\right)}{dx}=-co\sec h^{2}x$