Answer
$$\frac{{dy}}{{dt}} = 10{\left( {t\tan t} \right)^9}\left( {t{{\sec }^2}t + \tan t} \right)$$
Work Step by Step
$$\eqalign{
& y = {\left( {t\tan t} \right)^{10}} \cr
& {\text{differentiate with respect to }}t \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {{{\left( {t\tan t} \right)}^{10}}} \right] \cr
& {\text{use the chain rule }}\frac{d}{{dt}}\left[ {{u^n}} \right] = n{u^{n - 1}}\frac{{du}}{{dt}} \cr
& \frac{{dy}}{{dt}} = 10{\left( {t\tan t} \right)^9}\frac{d}{{dt}}\left[ {t\tan t} \right] \cr
& {\text{use the product rule}} \cr
& \frac{{dy}}{{dt}} = 10{\left( {t\tan t} \right)^9}\left( {t\frac{d}{{dt}}\left[ {\tan t} \right] + \tan t\frac{d}{{dt}}\left[ t \right]} \right) \cr
& {\text{solve derivatives and simplify}} \cr
& \frac{{dy}}{{dt}} = 10{\left( {t\tan t} \right)^9}\left( {t\left( {{{\sec }^2}t} \right) + \tan t\left( 1 \right)} \right) \cr
& \frac{{dy}}{{dt}} = 10{\left( {t\tan t} \right)^9}\left( {t{{\sec }^2}t + \tan t} \right) \cr} $$
