Answer
$${\text{false}}$$
Work Step by Step
$$\eqalign{
& {\text{calculate }}{D_x}\tan \left( {{x^2}} \right) \cr
& {\text{use the chain rule for }}{D_x}\tan x = {\sec ^2}x,{\text{ then }}{D_x}\tan u = {\sec ^2}u \cdot {D_x}u \cr
& then \cr
& {\text{ }}{D_x}\tan \left( {{x^2}} \right) = {\text{ se}}{{\text{c}}^2}\left( {{x^2}} \right) \cdot {D_x}\left( {{x^2}} \right) \cr
& {\text{ }}{D_x}\tan \left( {{x^2}} \right) = {\text{ se}}{{\text{c}}^2}\left( {{x^2}} \right)\left( {2x} \right) \cr
& {\text{ }}{D_x}\tan \left( {{x^2}} \right) = 2x{\text{ se}}{{\text{c}}^2}\left( {{x^2}} \right) \cr
& {\text{therefore}}{\text{, the satement }}{D_x}\tan \left( {{x^2}} \right) = {\text{ se}}{{\text{c}}^2}\left( {{x^2}} \right){\text{ is false}} \cr} $$
