Question

find $\frac{ds}{dt}$. $s = \frac{2+cot t}{2 - cot t}$ $\frac{ds}{dt}=square$

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $s=\frac{u}{v}$, then $\frac{ds}{dt}=\frac{u'v - uv'}{v^{2}}$. Here, $u = 2+\cot t$, $u'=-\csc^{2}t$, $v = 2-\cot t$, and $v'=\csc^{2}t$.

Step2: Substitute into the quotient - rule formula

$\frac{ds}{dt}=\frac{(-\csc^{2}t)(2 - \cot t)-(2+\cot t)(\csc^{2}t)}{(2 - \cot t)^{2}}$.

Step3: Expand the numerator

Expand the numerator: $(-\csc^{2}t)(2 - \cot t)-(2+\cot t)(\csc^{2}t)=-2\csc^{2}t+\csc^{2}t\cot t-2\csc^{2}t-\csc^{2}t\cot t$.

Step4: Simplify the numerator

Combine like - terms in the numerator: $-2\csc^{2}t+\csc^{2}t\cot t-2\csc^{2}t-\csc^{2}t\cot t=-4\csc^{2}t$.
So, $\frac{ds}{dt}=\frac{-4\csc^{2}t}{(2 - \cot t)^{2}}$.

Answer:

$\frac{-4\csc^{2}t}{(2 - \cot t)^{2}}$