Answer
$$\eqalign{
& h'\left( x \right) = {x^{\sqrt x }}\left( {\frac{{2 + \ln x}}{{2\sqrt x }}} \right) \cr
& h'\left( 4 \right) = 4\left( {2 + \ln 4} \right) \cr} $$
Work Step by Step
$$\eqalign{
& h\left( x \right) = {x^{\sqrt x }};\,\,\,\,\,\,a = 4 \cr
& {\text{Use }}{b^x} = {e^{x\ln b}},{\text{ then}} \cr
& h\left( x \right) = {e^{\sqrt x \ln x}} \cr
& {\text{Differentiating}} \cr
& h'\left( x \right) = {e^{\sqrt x \ln x}}\frac{d}{{dx}}\left[ {\sqrt x \ln x} \right] \cr
& h'\left( x \right) = {e^{\sqrt x \ln x}}\left[ {\sqrt x \left( {\frac{1}{x}} \right) + \frac{{\ln x}}{{2\sqrt x }}} \right] \cr
& h'\left( x \right) = {x^{\sqrt x }}\left[ {\frac{1}{{\sqrt x }} + \frac{{\ln x}}{{2\sqrt x }}} \right] \cr
& h'\left( x \right) = {x^{\sqrt x }}\left( {\frac{{2 + \ln x}}{{2\sqrt x }}} \right) \cr
& \cr
& {\text{Evaluate the derivative at }}x = 4 \cr
& h'\left( 4 \right) = {4^{\sqrt 4 }}\left( {\frac{{2 + \ln 4}}{{2\sqrt 4 }}} \right) \cr
& h'\left( 4 \right) = 16\left( {\frac{{2 + \ln 4}}{4}} \right) \cr
& h'\left( 4 \right) = 4\left( {2 + \ln 4} \right) \cr} $$
