Answer
$y' = \frac{{\left( {\ln 5} \right)\left( {{5^{\sqrt x }}} \right)}}{{2\sqrt x }}$
Work Step by Step
$$\eqalign{
& y = {5^{\sqrt x }} \cr
& {\text{Differentiate}} \cr
& y' = \frac{d}{{dx}}\left[ {{5^{\sqrt x }}} \right] \cr
& {\text{Let }}g\left( x \right) = \sqrt x ,{\text{ then}} \cr
& y' = \frac{d}{{dx}}\left[ {{5^{g\left( x \right)}}} \right] \cr
& {\text{By using the chain rule and }}\frac{d}{{dx}}\left[ {{b^x}} \right] = {b^x}\ln b{\text{ }} \cr
& y' = {5^{g\left( x \right)}}\left( {\ln 5} \right)g'\left( x \right) \cr
& {\text{Back - substitute }}g\left( x \right) = \sqrt x \cr
& y' = {5^{\sqrt x }}\left( {\ln 5} \right)\frac{d}{{dx}}\left[ {\sqrt x } \right] \cr
& y' = {5^{\sqrt x }}\left( {\ln 5} \right)\left( {\frac{1}{{2\sqrt x }}} \right) \cr
& y' = \frac{{\left( {\ln 5} \right)\left( {{5^{\sqrt x }}} \right)}}{{2\sqrt x }} \cr} $$
