What is the derivative of $y = (sec x + tan x) (sec x - tan x)$ $y = (secx + tanx)(secx -tanx)$ ?

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Answer 1 · 2016-12-07T17:49:27

$y' = 0$

Start by rewriting in terms of sine and cosine. For this problem we use the identities $sectheta = 1/costheta$ ad $tantheta = sintheta/costheta$ .

$y = (1/cosx + sinx/cosx)(1/cosx- sinx/cosx)$

$y = ((1 + sinx)/cosx)((1 - sinx)/cosx)$

$y = (1 - sin^2x)/cos^2x$

$y = cos^2x/cos^2x$

$y = 1$

The derivative of any constant is $0$ .

$y' = 0$

Hopefully this helps!

What is the derivative of y = (secx + tanx)(secx -tanx)y=(secx+tanx)(secx−tanx)y = (secx + tanx)(secx -tanx)?