Answer
$$y' = {\sec ^5}x{\tan ^3}x$$
Work Step by Step
$$\eqalign{
& y = \frac{{{{\sec }^7}x}}{7} - \frac{{{{\sec }^5}x}}{5} \cr
& {\text{Differentiating}} \cr
& y' = \frac{d}{{dx}}\left[ {\frac{{{{\sec }^7}x}}{7}} \right] - \frac{d}{{dx}}\left[ {\frac{{{{\sec }^5}x}}{5}} \right] \cr
& {\text{Pull out the constants}} \cr
& y' = \frac{1}{7}\frac{d}{{dx}}\left[ {{{\sec }^7}x} \right] - \frac{1}{5}\frac{d}{{dx}}\left[ {{{\sec }^5}x} \right] \cr
& {\text{compute derivatives using the chain rule}} \cr
& y' = \frac{1}{7}\left( {7{{\sec }^6}x} \right)\frac{d}{{dx}}\left[ {\sec x} \right] - \frac{1}{5}\left( {5{{\sec }^4}x} \right)\frac{d}{{dx}}\left[ {\sec x} \right] \cr
& y' = {\sec ^6}x\frac{d}{{dx}}\left[ {\sec x} \right] - {\sec ^4}x\frac{d}{{dx}}\left[ {\sec x} \right] \cr
& y' = {\sec ^6}x\left( {\sec x\tan x} \right) - {\sec ^4}x\left( {\sec x\tan x} \right) \cr
& y' = {\sec ^7}x\tan x - {\sec ^5}x\tan x \cr
& {\text{Factoring}} \cr
& y' = {\sec ^5}x\tan x\left( {{{\sec }^2}x - 1} \right) \cr
& y' = {\sec ^5}x\tan x\left( {{{\tan }^2}x} \right) \cr
& y' = {\sec ^5}x{\tan ^3}x \cr} $$
