How do you find the derivative of y=ln abs(sec x-tanx)?

Feb 21, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} - \sec x$

Explanation:

$\ln f \left(x\right) = \frac{f ' \left(x\right)}{f \left(x\right)}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sec x \tan x - {\sec}^{2} x}{\sec x - \tan x} =$

$\frac{- \left({\sec}^{2} x - \sec \tan x\right)}{\sec x - \tan x} =$

$\frac{- \sec x \left(\sec x - \tan x\right)}{\sec x - \tan x} =$

$- \sec x$