# What is the derivative of (cos x) / (1 + tan x)?

Apr 14, 2015

The answer is: $y ' = \frac{- \sin x \cos x - {\sin}^{2} x - 1}{\cos x {\left(1 + \tan x\right)}^{2}}$.

Remembering the division rule, that is:

$y = f \frac{x}{g} \left(x\right) \Rightarrow y ' = \frac{f ' \left(x\right) \cdot g \left(x\right) - f \left(x\right) \cdot g ' \left(x\right)}{g \left(x\right)} ^ 2$,

than:

$y ' = \frac{- \sin x \left(1 + \tan x\right) - \cos x \cdot \frac{1}{\cos} ^ 2 x}{1 + \tan x} ^ 2 =$

$= \frac{- \sin x - \sin x \cdot \sin \frac{x}{\cos} x - \frac{1}{\cos} x}{1 + \tan x} ^ 2 =$

$= \frac{- \sin x \cos x - {\sin}^{2} x - 1}{\cos x {\left(1 + \tan x\right)}^{2}}$.