# How do you use the quotient rule to differentiate csc(t)/tan(t) ?

Jun 6, 2017

$\frac{d}{\mathrm{dx}} \left(\csc \frac{t}{\tan} t\right) = - \csc x \left({\cot}^{2} x + {\csc}^{2} x\right)$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left(\frac{p \left(x\right)}{q \left(x\right)}\right) = \frac{q \left(x\right) p ' \left(x\right) - p \left(x\right) q ' \left(x\right)}{{\left[q \left(x\right)\right]}^{2}}$

let $p \left(x\right) = \csc t$

$p ' \left(x\right) = - \csc t \cot t$

let $q \left(x\right) = \tan t$

$q ' \left(x\right) = {\sec}^{2} t$

$\frac{d}{\mathrm{dx}} \left(\csc \frac{t}{\tan} t\right) = \frac{- \csc t \cot t \left(\tan t\right) - {\sec}^{2} t \csc t}{{\tan}^{2} t}$

$= - \csc x \left({\cot}^{2} x + {\csc}^{2} x\right)$