Answer
$$dy = {\sec ^2}xdx$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \tan x \cr
& {\text{Differentiate }} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\tan x} \right] \cr
& f'\left( x \right) = {\sec ^2}x \cr
& {\text{Write in the form }}dy = f'\left( x \right)dx \cr
& dy = {\sec ^2}xdx \cr} $$