Answer
$$h'\left( x \right) = {2^{{x^2} + 1}}x\left( {\ln 2} \right)$$
Work Step by Step
$$\eqalign{
& h\left( x \right) = {2^{\left( {{x^2}} \right)}} \cr
& {\text{evaluate the derivative}} \cr
& h'\left( x \right) = \frac{d}{{dx}}\left( {{2^{\left( {{x^2}} \right)}}} \right) \cr
& {\text{use }}\frac{d}{{dx}}\left( {{a^{u\left( x \right)}}} \right) = {a^{u\left( x \right)}}\left( {\ln a} \right)u'\left( x \right) \cr
& h'\left( x \right) = {2^{\left( {{x^2}} \right)}}\left( {\ln 2} \right)\frac{d}{{dx}}\left( {{x^2}} \right) \cr
& h'\left( x \right) = {2^{\left( {{x^2}} \right)}}\left( {\ln 2} \right)\left( {2x} \right) \cr
& {\text{simplify}} \cr
& h'\left( x \right) = {2^{{x^2} + 1}}x\left( {\ln 2} \right) \cr} $$
