Answer
$$\frac{{dy}}{{dx}} = {3^x}\left( {\ln 3} \right)$$
Work Step by Step
$$\eqalign{
& y = {3^x} \cr
& \left( i \right){\text{Using the fact that }}{b^x} = {e^{x\ln b}},{\text{ then}} \cr
& y = {e^{x\ln 3}} \cr
& {\text{Differentiating}} \cr
& \frac{{dy}}{{dx}} = {e^{x\ln 3}}\frac{d}{{dx}}\left[ {x\ln 3} \right] \cr
& \frac{{dy}}{{dx}} = {e^{x\ln 3}}\left( {\ln 3} \right) \cr
& {\text{Substitute }}{3^x}{\text{ for }}{e^{x\ln 3}} \cr
& \frac{{dy}}{{dx}} = {3^x}\left( {\ln 3} \right) \cr
& \cr
& \left( {ii} \right){\text{Using the logarithmic differentiation}} \cr
& \ln y = \ln {3^x} \cr
& \ln y = x\ln 3 \cr
& {\text{Differentiate both sides}} \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \ln 3 \cr
& {\text{Solving for }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = y\ln 3 \cr
& {\text{Where }}y = {3^x} \cr
& \frac{{dy}}{{dx}} = {3^x}\left( {\ln 3} \right) \cr} $$
