How do you differentiate given y = (sec)^2 x + (tan)^2 x?

Mar 2, 2018

$y ' = 4 {\sec}^{2} x \tan x$

Explanation:

$y = {\sec}^{2} x + {\tan}^{2} x$
$y = \sec x \sec x + \tan x \tan x$

$y ' = \frac{\mathrm{dy}}{\mathrm{dx}} = \left(\sec x \tan x\right) \sec x + \sec x \left(\sec x \tan x\right) + \left({\sec}^{2} x\right) \tan x + \tan x \left({\sec}^{2} x\right)$

$y ' = {\sec}^{2} x \tan x + {\sec}^{2} x \tan x + {\sec}^{2} x \tan x + {\sec}^{2} x \tan x$

$y ' = 4 {\sec}^{2} x \tan x$