Answer
$$y = {2^{{x^2} + 1}}x\ln 2$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {2^{{x^2}}} \cr
& or \cr
& y = {2^{{x^2}}} \cr
& {\text{take logarithm natural on both sides}} \cr
& \ln y = \ln {2^{{x^2}}} \cr
& {\text{logarithm properties}} \cr
& \ln y = {x^2}\ln 2 \cr
& {\text{differentiate}} \cr
& \left( {\ln y} \right)' = \left( {{x^2}\ln 2} \right)' \cr
& \left( {\ln y} \right)' = \ln 2\left( {{x^2}} \right)' \cr
& \frac{{y'}}{y} = \left( {\ln 2} \right)\left( {2x} \right) \cr
& y' = y\left( {\ln 2} \right)\left( {2x} \right) \cr
& {\text{replace }}y = {2^{{x^2}}} \cr
& y' = \left( {{2^{{x^2}}}} \right)\left( {\ln 2} \right)\left( {2x} \right) \cr
& {\text{simplify}} \cr
& y = {2^{{x^2} + 1}}x\ln 2 \cr} $$