Answer
$$dy = 2x\sec \left( {{x^2} - 1} \right)\tan \left( {{x^2} - 1} \right)dx$$
Work Step by Step
$$\eqalign{
& y = \sec \left( {{x^2} - 1} \right) \cr
& {\text{Calculate }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\sec \left( {{x^2} - 1} \right)} \right] \cr
& {\text{Recall that }}\frac{d}{{dx}}\left[ {\sec u} \right] = \sec u\tan u\frac{{du}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \sec \left( {{x^2} - 1} \right)\tan \left( {{x^2} - 1} \right)\frac{d}{{dx}}\left[ {{x^2} - 1} \right] \cr
& \frac{{dy}}{{dx}} = \sec \left( {{x^2} - 1} \right)\tan \left( {{x^2} - 1} \right)\left( {2x} \right) \cr
& \frac{{dy}}{{dx}} = 2x\sec \left( {{x^2} - 1} \right)\tan \left( {{x^2} - 1} \right) \cr
& {\text{Write in differential form}} \cr
& dy = 2x\sec \left( {{x^2} - 1} \right)\tan \left( {{x^2} - 1} \right)dx \cr} $$
