# Find the derivative of (x+cosx)/tanx?

Feb 7, 2018

Derivative is $\frac{\tan x - \sin x \tan x + x {\sec}^{2} x + \sec x}{\tan} ^ 2 x$

#### Explanation:

We use quotient rule, which says that derivative of $\frac{f \left(x\right)}{g \left(x\right)}$ is

$\frac{g \left(x\right) f ' \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$

Hence derivative of $\frac{x + \cos x}{\tan} x$ is

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

= $\frac{\tan x - \sin x \tan x + x {\sec}^{2} x + \sec x}{\tan} ^ 2 x$