Find the derivative of the following functions (it is to be understood that a \dfrac{cosx}{1+sinx}

Let f(x) = \( \dfrac{cosx}{1+sinx}\)

By quotient rule,

\(=\dfrac{(1+sinx)\dfrac{d}{dx}(cosx)-(cosx)\dfrac{d}{dx}(1+sinx)}{(1+sinx)^2}\)

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

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

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

= \( \dfrac{-(1-sinx)}{(1+sinx)^2} \)

= \(\dfrac{-1}{ (1+sinx)^2}\)