Answer
$$h'\left( z \right) = - z{e^{ - \frac{{{z^2}}}{2}}}$$
Work Step by Step
$$\eqalign{
& h\left( z \right) = {e^{ - \frac{{{z^2}}}{2}}} \cr
& {\text{Differentiate}} \cr
& h'\left( z \right) = \frac{d}{{dz}}\left[ {{e^{ - \frac{{{z^2}}}{2}}}} \right] \cr
& h'\left( z \right) = {e^{ - \frac{{{z^2}}}{2}}}\frac{d}{{dz}}\left[ { - \frac{{{z^2}}}{2}} \right] \cr
& h'\left( z \right) = {e^{ - \frac{{{z^2}}}{2}}}\left( {\frac{{ - 2z}}{2}} \right) \cr
& h'\left( z \right) = - z{e^{ - \frac{{{z^2}}}{2}}} \cr} $$