Question

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

Answer 1

`dy/dx-secx`

Explanation:

`lnf(x)=(f'(x))/(f(x))`

`dy/dx=(secxtanx-sec^2x)/(secx-tanx)=`

`(-(sec^2x-sectanx))/(secx-tanx)=`

`(-secx(secx-tanx))/(secx-tanx)=`

`-secx`