Find the derivative of the following functions (it is to be understood that \dfrac{secx-1}{secx+1}

Let f(x) = ​​\( \dfrac{secx-1}{secx+1}\)

\( f(x)=\dfrac{\dfrac{1}{cosx}-1}{\dfrac{1}{cosx}+1}=\dfrac{1-cosx}{1+cosx}\)

By quotient rule,

= \( \dfrac{sinx+cosxsinx+sinx-sinxcosx}{(1+cos^2x)}\)

= \( \dfrac{2sinx}{(1+cos^2x)}\)

\( \dfrac{2sinx}{(\dfrac{1}{1+secx})^2}=\dfrac{2sinx}{\dfrac{(secx+1)^2}{sec^2x}}\)

= \( \dfrac{2sinxsec^2x}{(secx+1)^2}\)

= \( \dfrac{\dfrac{2sinx}{cosx}secx}{(secx+1)^2}\)

= \( \dfrac{2secxtanx}{(secx+1)}\)