Differentiate the following functions with respect to `x :` `(secx-1)/(secx+1)`

Explanation

`frac{d}{d x}(frac{sec x-1}{sec x+1})`



we have `frac{d(frac{u}{v})}{d x}=frac{v frac{d u}{d x}-u frac{d v}{d x}}{v^{2}}`



`=frac{(sec x+1) frac{d}{d x}(sec x-1)-(sec x-1) frac{d}{d x}(sec x+1)}{(sec x+1)^{2}}`




`=frac{(sec x+1) sec x tan x-(sec x-1)(sec x tan x)}{(sec x+1)^{2}}`


...