Answer
$$f'\left( x \right) = 4\left( {{3^x}} \right)\left( {\frac{{x\left( {\ln 3} \right) - 1/2}}{{x\sqrt x }}} \right)$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{4\left( {{3^x}} \right)}}{{\sqrt x }} \cr
& {\text{calculate the derivative of the function}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left( {\frac{{4\left( {{3^x}} \right)}}{{\sqrt x }}} \right) \cr
& f'\left( x \right) = 4\frac{d}{{dx}}\left( {\frac{{{3^x}}}{{\sqrt x }}} \right) \cr
& {\text{Use the quotient rule }}\left( {{\text{see the page 237}}} \right) \cr
& f'\left( x \right) = 4\left( {\frac{{\sqrt x \left( {{3^x}} \right)' - {3^x}\left( {\sqrt x } \right)'}}{{{{\left( {\sqrt x } \right)}^2}}}} \right) \cr
& {\text{simplifying}} \cr
& f'\left( x \right) = 4\left( {\frac{{\sqrt x \left( {{3^x}\ln 3} \right) - {3^x}\left( {1/2\sqrt x } \right)}}{x}} \right) \cr
& f'\left( x \right) = \frac{{4\left( {{3^x}} \right)}}{{\sqrt x }}\left( {\frac{{x\left( {\ln 3} \right) - 1/2}}{x}} \right) \cr
& f'\left( x \right) = 4\left( {{3^x}} \right)\left( {\frac{{x\left( {\ln 3} \right) - 1/2}}{{x\sqrt x }}} \right) \cr} $$
