# How do you find the derivative of  ( cos (x) ) / ( 2 + sin (x) )?

Sep 26, 2015

$\frac{- 2 \sin x - 1}{2 + \sin x} ^ 2$

#### Explanation:

$f \left(x\right) = \cos \frac{x}{2 + \sin x}$
let $u \left(x\right) = \cos \left(x\right)$ and $v \left(x\right) = 2 + \sin x$
then $u ' \left(x\right) = - \sin x$ and $v ' \left(x\right) = \cos x$
$f \left(x\right) = \frac{u \left(x\right)}{v \left(x\right)}$
$f ' \left(x\right) = \frac{1}{v {\left(x\right)}^{2}} \left[v \left(x\right) u ' \left(x\right) - u \left(x\right) v ' \left(x\right)\right]$
$f ' \left(x\right) = \frac{1}{2 + \sin x} ^ 2 \left[\left(2 + \sin x\right) \left(- \sin x\right) - \cos x \left(\cos x\right)\right] = \frac{- 2 \sin x - 1}{2 + \sin x} ^ 2$