Answer
$$f'\left( w \right) = - \frac{{4w}}{{\sqrt {1 - 4{w^2}} }}$$
Work Step by Step
$$\eqalign{
& f\left( w \right) = \cos \left( {{{\sin }^{ - 1}}2w} \right) \cr
& {\text{find the derivative}}{\text{, differentiate both sides with respect to }}w \cr
& f'\left( w \right) = \frac{d}{{dw}}\left[ {\cos \left( {{{\sin }^{ - 1}}2w} \right)} \right] \cr
& {\text{use }}\frac{d}{{dw}}\left[ {\cos u} \right] = - \sin u\frac{{du}}{{dw}}.{\text{ let }}u = {\sin ^{ - 1}}2w \cr
& f'\left( w \right) = - \sin \left( {{{\sin }^{ - 1}}2w} \right)\frac{d}{{dw}}\left[ {{{\sin }^{ - 1}}2w} \right] \cr
& {\text{use }}\frac{d}{{dw}}\left[ {{{\sin }^{ - 1}}u} \right] = \frac{1}{{\sqrt {1 - {u^2}} }}\frac{{du}}{{dw}}.{\text{ let }}u = 2w \cr
& f'\left( w \right) = - \sin \left( {{{\sin }^{ - 1}}2w} \right)\left( {\frac{1}{{\sqrt {1 - {{\left( {2w} \right)}^2}} }}} \right)\frac{d}{{dw}}\left[ {2w} \right] \cr
& f'\left( w \right) = - \sin \left( {{{\sin }^{ - 1}}2w} \right)\left( {\frac{1}{{\sqrt {1 - {{\left( {2w} \right)}^2}} }}} \right)\left( 2 \right) \cr
& {\text{simplifying}} \cr
& f'\left( w \right) = - \left( {2w} \right)\left( {\frac{1}{{\sqrt {1 - 4{w^2}} }}} \right)\left( 2 \right) \cr
& f'\left( w \right) = - \frac{{4w}}{{\sqrt {1 - 4{w^2}} }} \cr} $$
