Answer
$G'\left( z \right) = - 6\sin z\cos z{\left( {1 + {{\cos }^2}z} \right)^2}$
Work Step by Step
$$\eqalign{
& G\left( z \right) = {\left( {1 + {{\cos }^2}z} \right)^3} \cr
& {\text{Differentiate}} \cr
& G'\left( z \right) = \frac{d}{{dz}}\left[ {{{\left( {1 + {{\cos }^2}z} \right)}^3}} \right] \cr
& {\text{Use The Power Rule Combined with the Chain Rule}} \cr
& G'\left( z \right) = 3{\left( {1 + {{\cos }^2}z} \right)^{3 - 1}}\frac{d}{{dz}}\left[ {1 + {{\cos }^2}z} \right] \cr
& G'\left( z \right) = 3{\left( {1 + {{\cos }^2}z} \right)^2}\left[ {\frac{d}{{dz}}\left[ 1 \right] + \frac{d}{{dz}}\left[ {{{\cos }^2}z} \right]} \right] \cr
& {\text{Apply the chain rule }} \cr
& G'\left( z \right) = 3{\left( {1 + {{\cos }^2}z} \right)^2}\left[ {\frac{d}{{dz}}\left[ 1 \right] + 2\cos z\frac{d}{{dz}}\left[ {\cos z} \right]} \right] \cr
& {\text{Compute derivatives and simplify}} \cr
& G'\left( z \right) = 3{\left( {1 + {{\cos }^2}z} \right)^2}\left[ {0 + 2\cos z\left( { - \sin z} \right)} \right] \cr
& G'\left( z \right) = 3{\left( {1 + {{\cos }^2}z} \right)^2}\left( { - 2\sin z\cos z} \right) \cr
& G'\left( z \right) = - 6\sin z\cos z{\left( {1 + {{\cos }^2}z} \right)^2} \cr} $$
