How do you differentiate  f(x) =(sin x + tan x)/(sin x-cos x) ?

Dec 21, 2015

$f ' \left(x\right) = \frac{\left(\cos x - {\sec}^{2} x\right) \left(\sin x - \cos x\right) - \left(\cos x + \sin x\right) \left(\sin x + \tan x\right)}{\sin x - \cos x} ^ 2$

Explanation:

Original equation:
$f \left(x\right) = \frac{\sin x + \tan x}{\sin x - \cos x}$

Use the quotient rule to derive.

Derive the top and times by the bottom:
$\left(\cos x - {\sec}^{2} x\right) \left(\sin x - \cos x\right)$

Derive the bottom and multiply by the top:
$\left(\cos x + \sin x\right) \left(\sin x + \tan x\right)$

Subtract the two:
$\left(\cos x - {\sec}^{2} x\right) \left(\sin x - \cos x\right) - \left(\cos x + \sin x\right) \left(\sin x + \tan x\right)$

Place it over the bottom squared:
$\frac{\left(\cos x - {\sec}^{2} x\right) \left(\sin x - \cos x\right) - \left(\cos x + \sin x\right) \left(\sin x + \tan x\right)}{\sin x - \cos x} ^ 2$

$f ' \left(x\right) = \frac{\left(\cos x - {\sec}^{2} x\right) \left(\sin x - \cos x\right) - \left(\cos x + \sin x\right) \left(\sin x + \tan x\right)}{\sin x - \cos x} ^ 2$