Answer
$$\frac{{dy}}{{dx}} = \frac{5}{{\left| x \right|\sqrt {{x^{10}} - 1} }}$$
Work Step by Step
$$\eqalign{
& y = {\sec ^{ - 1}}\left( {{x^5}} \right) \cr
& {\text{differentiate using the formula }} \cr
& \frac{d}{{dx}}\left[ {{{\sec }^{ - 1}}u} \right] = \frac{1}{{\left| u \right|\sqrt {{u^2} - 1} }}\frac{{du}}{{dx}}{\text{ }}\left( {{\text{see page 467}}} \right) \cr
& u = {x^5} \cr
& \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\left| {{x^5}} \right|\sqrt {{{\left( {{x^5}} \right)}^2} - 1} }}\frac{d}{{dx}}\left( {{x^5}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\left| {{x^5}} \right|\sqrt {{{\left( {{x^5}} \right)}^2} - 1} }}\left( {5{x^4}} \right) \cr
& {\text{simplifying}}{\text{, }} \cr
& \frac{{dy}}{{dx}} = \frac{{5{x^4}}}{{\left| {{x^5}} \right|\sqrt {{x^{10}} - 1} }} \cr
& \frac{{dy}}{{dx}} = \frac{{5{x^4}}}{{{x^4}\left| x \right|\sqrt {{x^{10}} - 1} }} \cr
& \frac{{dy}}{{dx}} = \frac{5}{{\left| x \right|\sqrt {{x^{10}} - 1} }} \cr} $$