Find the derivative of the functions sec x

Let f(x) =  sec x

= \( \dfrac{1}{cosx}\)

By differentiating both sides, we get

\( f'(x) =\dfrac{d}{dx}(\dfrac{1}{cosx})\)

\( f'(x) =cosx\dfrac{cosx\dfrac{d}{dx}(1)-1\dfrac{d}{dx}(cosx)}{cos^2x}\)

\( \dfrac{cosx\times 0-(-sinx)}{cos^2x}\)

We get

= \( \dfrac{sinx}{cos^2x}= \dfrac{sinx}{cosx}\times \dfrac{1}{cosx}\)

= tan x sec x