#### Answer

The Derivative is:
$\frac{ds}{dt}=\frac{-2\csc t\cot t}{(1-\csc t)^2}$

#### Work Step by Step

$s=\frac{1\space+\space\csc t}{1\space-\space\csc t}$
Applying Derivative Rules:
$y'=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{g^2(x)}$
$\frac{ds}{dt}=\frac{(1-\csc t)\cdot\frac{d}{dt}(1+\csc t)-(1+\csc t)\cdot\frac{d}{dt}(1-\csc t)}{(1-\csc t)^2}$
$\frac{ds}{dt}=\frac{(1-\csc t)(0+(-\csc t\cot t))-(1+\csc t)(0-(-\csc t\cot t))}{(1-\csc t)^2}$
$\frac{ds}{dt}=\frac{-\csc t\cot t+\csc^2 t\cot t-\csc t\cot t-\csc^2t\cot t}{(1-\csc t)^2}$
$\frac{ds}{dt}=\frac{-2\csc t\cot t}{(1-\csc t)^2}$