Answer
$f'\left( w \right) = \frac{{2\sec w\tan w}}{{{{\left( {1 - \sec w} \right)}^2}}}$
Work Step by Step
$$\eqalign{
& f\left( w \right) = \frac{{1 + \sec w}}{{1 - \sec w}} \cr
& {\text{Differentiating}} \cr
& f'\left( w \right) = \frac{d}{{dw}}\left[ {\frac{{1 + \sec w}}{{1 - \sec w}}} \right] \cr
& {\text{Use the quotient rule for derivatives }}\left( {\frac{g}{h}} \right)' = \frac{{hg' - gh'}}{{{h^2}}} \cr
& {\text{Let }}g = 1 + \sec w{\text{ and }}h = 1 - \sec w,{\text{ then}} \cr
& f'\left( w \right) = \frac{{\left( {1 - \sec w} \right)\left( {1 + \sec w} \right)' - \left( {1 + \sec w} \right)\left( {1 - \sec w} \right)'}}{{{{\left( {1 - \sec w} \right)}^2}}} \cr
& {\text{Use the Derivatives of Trigonometric Functions }} \cr
& f'\left( w \right) = \frac{{\left( {1 - \sec w} \right)\left( {\sec w\tan w} \right) - \left( {1 + \sec w} \right)\left( { - \sec w\tan w} \right)}}{{{{\left( {1 - \sec w} \right)}^2}}} \cr
& f'\left( w \right) = \frac{{\left( {1 - \sec w} \right)\left( {\sec w\tan w} \right) + \left( {1 + \sec w} \right)\left( {\sec w\tan w} \right)}}{{{{\left( {1 - \sec w} \right)}^2}}} \cr
& {\text{Multiply and simplify}} \cr
& f'\left( w \right) = \frac{{\sec w\tan w - {{\sec }^2}w\tan w + \sec w\tan w + {{\sec }^2}w\tan w}}{{{{\left( {1 - \sec w} \right)}^2}}} \cr
& f'\left( w \right) = \frac{{\sec w\tan w + \sec w\tan w}}{{{{\left( {1 - \sec w} \right)}^2}}} \cr
& f'\left( w \right) = \frac{{2\sec w\tan w}}{{{{\left( {1 - \sec w} \right)}^2}}} \cr} $$
