Answer
$$g'\left( x \right) = \sin x + \tan x\sec x$$
Work Step by Step
$$\eqalign{
& g\left( x \right) = \left( {\sin x} \right)\left( {\tan x} \right) \cr
& {\text{Differentiate}} \cr
& g'\left( x \right) = \frac{d}{{dx}}\left[ {\left( {\sin x} \right)\left( {\tan x} \right)} \right] \cr
& {\text{By the product rule}} \cr
& g'\left( x \right) = \tan x\frac{d}{{dx}}\left[ {\sin x} \right] + \sin x\frac{d}{{dx}}\left[ {\tan x} \right] \cr
& g'\left( x \right) = \tan x\left( {\cos x} \right) + \sin x\left( {{{\sec }^2}x} \right) \cr
& {\text{Simplifying}} \cr
& g'\left( x \right) = \left( {\frac{{\sin x}}{{\cos x}}} \right)\left( {\cos x} \right) + \sin x\left( {\frac{1}{{{{\cos }^2}x}}} \right) \cr
& g'\left( x \right) = \sin x + \tan x\sec x \cr} $$
