Find dy/dx Calculus help?

Question

y = 3 (tan x + sec x)(tan x - sec x)

Answer 1

Given that y = 3 (tan x + sec x)(tan x - sec x)

                    = 3(tan^2x - sec^2x)                                     [ Since (a + b) (a - b) = a^2 - b^2 ]

                    = 3tan^2x - 3sec^2x

Differenciating on both sides we get,

             dy/dx = d/dx(3tan^2x) - d/dx(3sec^2x)

                      = 3(2tanx)d/dx(tanx) - 3(2secx)d/dx(secx)  [ Using chain rule ]

                      = 6tanx(sec^2x) - 6secx(secxtanx)   [Since d/dx(tanx) = sec^2x, d/dx(secx) = secxtanx]

                      = 6tanxsec^2x - 6tanxsec^2x

                      = 0

Therefore dy/dx = 0