Answer
$y' = \frac{{\tan x}}{{\ln 10}}$
Work Step by Step
$$\eqalign{
& y = {\log _{10}}\sec x \cr
& {\text{Differentiate}} \cr
& y' = \frac{d}{{dx}}\left[ {{{\log }_{10}}\sec x} \right] \cr
& {\text{Use the formula }}\frac{d}{{dx}}\left( {{{\log }_b}x} \right) = \frac{1}{{x\ln b}}{\text{ and the chain rule}} \cr
& y' = \frac{1}{{\left( {\sec x} \right)\left( {\ln 10} \right)}}\frac{d}{{dx}}\left[ {\sec x} \right] \cr
& {\text{compute the derivative and simplify}} \cr
& y' = \frac{1}{{\left( {\sec x} \right)\left( {\ln 10} \right)}}\left( {\sec x\tan x} \right) \cr
& y' = \frac{{\tan x}}{{\ln 10}} \cr} $$
