Answer
$$\frac{{dy}}{{dx}} = 10{\sec ^2}5x$$
Work Step by Step
$$\eqalign{
& y = 2\tan 5x \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {2\tan 5x} \right] \cr
& {\text{use multiple constant rule}} \cr
& \frac{{dy}}{{dx}} = 2\frac{d}{{dx}}\left[ {\tan 5x} \right] \cr
& {\text{using the chain rule for }}{D_x}\left( {\tan u} \right) = {\sec ^2}u \cdot {D_x}\left( u \right).{\text{ then}} \cr
& \frac{{dy}}{{dx}} = 2\left( {{{\sec }^2}5x} \right)\frac{d}{{dx}}\left[ {5x} \right] \cr
& \frac{{dy}}{{dx}} = 2\left( {{{\sec }^2}5x} \right)\left( 5 \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = 10{\sec ^2}5x \cr} $$