Answer
$F'\left( t \right) = - \frac{8}{{{{\left( {2t + 1} \right)}^5}}}$
Work Step by Step
$$\eqalign{
& F\left( t \right) = {\left( {\frac{1}{{2t + 1}}} \right)^4} \cr
& {\text{Rewrite the function}}{\text{, recall that }}\frac{1}{{{u^n}}} = {u^{ - n}},{\text{ let }}u = 2t + 1 \cr
& F\left( t \right) = {\left( {{{\left( {2t + 1} \right)}^{ - 1}}} \right)^4} \cr
& F\left( t \right) = {\left( {2t + 1} \right)^{ - 4}} \cr
& {\text{Differentiate}} \cr
& F'\left( t \right) = \frac{d}{{dt}}\left[ {{{\left( {2t + 1} \right)}^{ - 4}}} \right] \cr
& {\text{Use The Power Rule Combined with the Chain Rule}} \cr
& \frac{d}{{dx}}{\left[ {g\left( x \right)} \right]^n} = n{\left[ {g\left( x \right)} \right]^{n - 1}}g'\left( x \right) \cr
& {\text{Therefore}}{\text{,}} \cr
& F'\left( t \right) = - 4{\left( {2t + 1} \right)^{ - 4 - 1}}\frac{d}{{dt}}\left[ {2t + 1} \right] \cr
& {\text{Computing derivatives}} \cr
& F'\left( t \right) = - 4{\left( {2t + 1} \right)^{ - 5}}\left( 2 \right) \cr
& {\text{Multiply and simplify}} \cr
& F'\left( t \right) = - 8{\left( {2t + 1} \right)^{ - 5}} \cr
& F'\left( t \right) = - \frac{8}{{{{\left( {2t + 1} \right)}^5}}} \cr} $$
