Answer
$$\frac{{dw}}{{dz}} = 4{z^3}{\text{ and }}\frac{{{d^2}w}}{{d{z^2}}} = 12{z^2}$$
Work Step by Step
$$\eqalign{
& w = \left( {z + 1} \right)\left( {z - 1} \right)\left( {{z^2} + 1} \right) \cr
& {\text{Use the distributive property}} \cr
& w = \left( {{z^2} - 1} \right)\left( {{z^2} + 1} \right) \cr
& w = {\left( {{z^2}} \right)^2} - {\left( { - 1} \right)^2} \cr
& w = {z^4} - 1 \cr
& \cr
& {\text{Find the first derivative}} \cr
& \frac{{dw}}{{dz}} = \frac{d}{{dz}}\left[ {{z^4} - 1} \right] \cr
& {\text{use }}\frac{d}{{dz}}\left[ {{z^n}} \right] = n{z^{n - 1}}{\text{ and }}\frac{d}{{dz}}\left[ c \right] = 0,{\text{ }}c{\text{ is a constant}} \cr
& \frac{{dw}}{{dz}} = 4{z^3} - 0 \cr
& \frac{{dw}}{{dz}} = 4{z^3} \cr
& \cr
& {\text{Find the second derivative}} \cr
& \frac{{{d^2}w}}{{d{z^2}}} = \frac{d}{{dz}}\left[ {4{z^3}} \right] \cr
& {\text{use }}\frac{d}{{dz}}\left[ {{z^n}} \right] = n{z^{n - 1}}{\text{ and }}\frac{d}{{dz}}\left[ c \right] = 0,{\text{ }}c{\text{ is a constant}} \cr
& \frac{{{d^2}w}}{{d{z^2}}} = 4\left( {3{z^2}} \right) \cr
& \frac{{{d^2}w}}{{d{z^2}}} = 12{z^2} \cr} $$
