Answer
$$u'\left( x \right) = 2x \cdot {4^{{x^2} - 1}}\left[ {\left( {{x^2} + 1} \right)\left( {\ln 4} \right) + 1} \right]$$
Work Step by Step
$$\eqalign{
& u\left( x \right) = \left( {{x^2} + 1} \right) \cdot {4^{{x^2} - 1}} \cr
& {\text{Differentiate}} \cr
& u'\left( x \right) = \frac{d}{{dx}}\left[ {\left( {{x^2} + 1} \right) \cdot {4^{{x^2} - 1}}} \right] \cr
& {\text{Use the product rule}} \cr
& u'\left( x \right) = \left( {{x^2} + 1} \right)\frac{d}{{dx}}\left[ {{4^{{x^2} - 1}}} \right] + {4^{{x^2} - 1}}\frac{d}{{dx}}\left[ {\left( {{x^2} + 1} \right)} \right] \cr
& {\text{Use the rule }}\frac{d}{{dx}}\left[ {{a^u}} \right] = {a^u}\left( {\ln a} \right)u' \cr
& u'\left( x \right) = \left( {{x^2} + 1} \right)\left( {{4^{{x^2} - 1}}} \right)\left( {\ln 4} \right)\frac{d}{{dx}}\left[ {{x^2} - 1} \right] + {4^{{x^2} - 1}}\frac{d}{{dx}}\left[ {\left( {{x^2} + 1} \right)} \right] \cr
& {\text{Compute derivatives and simplify}} \cr
& u'\left( x \right) = \left( {{x^2} + 1} \right)\left( {{4^{{x^2} - 1}}} \right)\left( {\ln 4} \right)\left( {2x} \right) + {4^{{x^2} - 1}}\left( {2x} \right) \cr
& u'\left( x \right) = 2x \cdot {4^{{x^2} - 1}}\left[ {\left( {{x^2} + 1} \right)\left( {\ln 4} \right) + 1} \right] \cr} $$
