Answer
$$f'\left( x \right) = - \frac{{2x{{\left( {{{\tan }^{ - 1}}\left( {{x^2} + 4} \right)} \right)}^{ - 2}}}}{{1 + {{\left( {{x^2} + 4} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{{{{\tan }^{ - 1}}\left( {{x^2} + 4} \right)}} \cr
& {\text{rewrite the function}} \cr
& f\left( x \right) = {\left[ {{{\tan }^{ - 1}}\left( {{x^2} + 4} \right)} \right]^{ - 1}} \cr
& {\text{differentiate both sides }} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left( {{{\left[ {{{\tan }^{ - 1}}\left( {{x^2} + 4} \right)} \right]}^{ - 1}}} \right) \cr
& {\text{using the chain rule}} \cr
& f'\left( x \right) = - {\left[ {{{\tan }^{ - 1}}\left( {{x^2} + 4} \right)} \right]^{ - 2}}\frac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( {{x^2} + 4} \right)} \right) \cr
& {\text{use }}\frac{d}{{du}}\left[ {{{\tan }^{ - 1}}u} \right] = \frac{1}{{1 + {u^2}}}\frac{{du}}{{dx}}.{\text{ let }}u = {x^2} + 4 \cr
& {\text{then}} \cr
& f'\left( x \right) = - {\left[ {{{\tan }^{ - 1}}\left( {{x^2} + 4} \right)} \right]^{ - 2}}\left( {\frac{1}{{1 + {{\left( {{x^2} + 4} \right)}^2}}}} \right)\frac{d}{{dx}}\left[ {{x^2} + 4} \right] \cr
& f'\left( x \right) = - {\left[ {{{\tan }^{ - 1}}\left( {{x^2} + 4} \right)} \right]^{ - 2}}\left( {\frac{1}{{1 + {{\left( {{x^2} + 4} \right)}^2}}}} \right)\left( {2x} \right) \cr
& {\text{simplifying}} \cr
& f'\left( x \right) = - \frac{{2x{{\left( {{{\tan }^{ - 1}}\left( {{x^2} + 4} \right)} \right)}^{ - 2}}}}{{1 + {{\left( {{x^2} + 4} \right)}^2}}} \cr} $$
