Answer
$$f'\left( u \right) = - \frac{1}{{\left| {2u + 1} \right|\sqrt {{u^2} + u} }}$$
Work Step by Step
$$\eqalign{
& f\left( u \right) = {\csc ^{ - 1}}\left( {2u + 1} \right) \cr
& {\text{find the derivative}} \cr
& f'\left( u \right) = \frac{d}{{du}}\left[ {{{\csc }^{ - 1}}\left( {2u + 1} \right)} \right] \cr
& {\text{use }}\frac{d}{{du}}\left[ {{{\csc }^{ - 1}}z} \right] = - \frac{1}{{\left| z \right|\sqrt {{z^2} - 1} }}\frac{{dz}}{{du}}.{\text{ let }}z = 2u + 1 \cr
& f'\left( u \right) = - \frac{1}{{\left| {2u + 1} \right|\sqrt {{{\left( {2u + 1} \right)}^2} - 1} }}\frac{d}{{du}}\left[ {2u + 1} \right] \cr
& {\text{then}} \cr
& f'\left( u \right) = - \frac{1}{{\left| {2u + 1} \right|\sqrt {{{\left( {2u + 1} \right)}^2} - 1} }}\left( 2 \right) \cr
& {\text{simplifying}} \cr
& f'\left( u \right) = - \frac{2}{{\left| {2u + 1} \right|\sqrt {{{\left( {2u + 1} \right)}^2} - 1} }} \cr
& f'\left( u \right) = - \frac{2}{{\left| {2u + 1} \right|\sqrt {4{u^2} + 4u + 1 - 1} }} \cr
& f'\left( u \right) = - \frac{2}{{\left| {2u + 1} \right|\sqrt {4{u^2} + 4u} }} \cr
& f'\left( u \right) = - \frac{2}{{2\left| {2u + 1} \right|\sqrt {{u^2} + u} }} \cr
& f'\left( u \right) = - \frac{1}{{\left| {2u + 1} \right|\sqrt {{u^2} + u} }} \cr} $$
