How do you find the derivative of 1/(sec x - tan x)?

Jul 30, 2016

$\frac{1}{1 - \sin x}$

Explanation:

First, simplify the function:

$y = \frac{1}{\frac{1}{\cos} x - \sin \frac{x}{\cos} x} = \frac{1}{\frac{1 - \sin x}{\cos} x} = \cos \frac{x}{1 - \sin x}$

From here, use the quotient rule.

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

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

${y}^{'} = \frac{- \sin x + {\sin}^{2} x + {\cos}^{2} x}{1 - \sin x} ^ 2$

Recall that ${\sin}^{2} x + {\cos}^{2} x = 1$:

${y}^{'} = \frac{1 - \sin x}{1 - \sin x} ^ 2$

${y}^{'} = \frac{1}{1 - \sin x}$

Note that there are many different ways this can be written, but this is the simplest.