# Find the derivative of x+cosx÷tanx with respect to x?

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Feb 25, 2017

 d/dx (x+cosx ÷ tanx ) = 1 - cosx - cotxcscx

#### Explanation:

 d/dx (x+cosx ÷ tanx ) = d/dx (x+cosx/tanx )
$\text{ } = \frac{d}{\mathrm{dx}} \left(x\right) + \frac{d}{\mathrm{dx}} \left(\cos \frac{x}{\tan} x\right)$
$\text{ } = 1 + \frac{\left(\tan x\right) \left(- \sin x\right) - \left(\cos x\right) \left({\sec}^{2} x\right)}{\tan x} ^ 2$
$\text{ } = 1 - \left(\tan x\right) \frac{\sin x}{\tan} ^ 2 x - \left(\cos x\right) \frac{{\sec}^{2} x}{\tan x} ^ 2$
$\text{ } = 1 - \left(\sin x\right) \cdot \cos \frac{x}{\sin} x - \left(\cos x\right) \left(\frac{1}{\cos} ^ 2 x\right) \cdot {\cos}^{2} \frac{x}{\sin} ^ 2 x$

$\text{ } = 1 - \cos x - \frac{\cos x}{\sin} ^ 2 x$

$\text{ } = 1 - \cos x - \cot x \csc x$

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