Answer
$G'\left( z \right) = \left( {1 - 4z} \right){\left( {{z^2} + 1} \right)^{ - 1/2}}\left( {z - 12{z^2} - 8} \right)$
Work Step by Step
$$\eqalign{
& G\left( z \right) = {\left( {1 - 4z} \right)^2}\sqrt {{z^2} + 1} \cr
& {\text{Differentiate}} \cr
& G'\left( z \right) = \frac{d}{{dz}}\left[ {{{\left( {1 - 4z} \right)}^2}\sqrt {{z^2} + 1} } \right] \cr
& {\text{Use the product rule for derivatives}} \cr
& G'\left( z \right) = {\left( {1 - 4z} \right)^2}\frac{d}{{dz}}\left[ {\sqrt {{z^2} + 1} } \right] + \sqrt {{z^2} + 1} \frac{d}{{dz}}\left[ {{{\left( {1 - 4z} \right)}^2}} \right] \cr
& {\text{Use the chain rule}} \cr
& G'\left( z \right) = {\left( {1 - 4z} \right)^2}\frac{1}{2}{\left( {{z^2} + 1} \right)^{ - 1/2}}\frac{d}{{dz}}\left[ {{z^2} + 1} \right] \cr
& + 2\sqrt {{z^2} + 1} \left( {1 - 4z} \right)\frac{d}{{dz}}\left[ {1 - 4z} \right] \cr
& {\text{Computing derivatives}} \cr
& G'\left( z \right) = {\left( {1 - 4z} \right)^2}\frac{1}{2}{\left( {{z^2} + 1} \right)^{ - 1/2}}\left( {2z} \right) + 2\sqrt {{z^2} + 1} \left( {1 - 4z} \right)\left( { - 4} \right) \cr
& {\text{Multiply}} \cr
& G'\left( z \right) = z{\left( {1 - 4z} \right)^2}{\left( {{z^2} + 1} \right)^{ - 1/2}} - 8\sqrt {{z^2} + 1} \left( {1 - 4z} \right) \cr
& {\text{Factor and simplify}} \cr
& G'\left( z \right) = \left( {1 - 4z} \right){\left( {{z^2} + 1} \right)^{ - 1/2}}\left[ {z\left( {1 - 4z} \right) - 8\left( {{z^2} + 1} \right)} \right] \cr
& G'\left( z \right) = \left( {1 - 4z} \right){\left( {{z^2} + 1} \right)^{ - 1/2}}\left( {z - 4{z^2} - 8{z^2} - 8} \right) \cr
& G'\left( z \right) = \left( {1 - 4z} \right){\left( {{z^2} + 1} \right)^{ - 1/2}}\left( {z - 12{z^2} - 8} \right) \cr} $$
