What is the derivative of ${sec}^{2} x - 1$ $sec^2 x - 1$ ?

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Answer 1 · 2016-09-28T09:31:45

$\frac{d}{d x} ({sec}^{2} x - 1) = 2 {sec}^{2} x tan x$ $d/(dx)(sec^2x-1)=2sec^2xtanx$

We can do it two ways.

One – as $\frac{d}{d x} sec x = sec x tan x$ $d/(dx)secx=secxtanx$

$\frac{d}{d x} ({sec}^{2} x - 1) = 2 sec x \times sec x tan x$ $d/(dx)(sec^2x-1)=2secx xx secxtanx$

= $2 {sec}^{2} x tan x$ $2sec^2xtanx$

Two – $\frac{d}{d x} ({sec}^{2} x - 1)$ $d/(dx)(sec^2x-1)$

= $\frac{d}{d x} {tan}^{2} x$ $d/(dx)tan^2x$

= $2 tan x \times {sec}^{2} x$ $2tanx xx sec^2x$

= $2 {sec}^{2} x tan x$ $2sec^2xtanx$

What is the derivative of sec2x−1sec^2 x - 1?