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Hello, Rose!

Quote:

Show that the derivative of $y\:=\:\frac{\sec x-\tan x}{\sec x+\tan x}$ .is: . $\frac{dy}{dx} \:=\: \frac{-2\sec x}{(\sec x+\tan x)^2}$

We can differentiate this head-on . . .

Quotient Rule:

. . $\frac{dy}{dx} \;=\;\frac{(\sec x + \tan x)(\sec x\tan x - \sec^2\!x) - (\sec x -\tan x)(\sec x\tan x + \sec^2\!x)}{(\sec x +\tan x)^2}$

. . . . $= \;\frac{-\sec x(\sec x + \tan x)(\sec x - \tan x) - \sec x(\sec x - \tan x)(\sec x + \tan x)}{(\sec x +\tan x)^2}$

. . . . $= \;\frac{-2\sec x(\sec x + \tan x)(\sec x -\tan x)}{(\sec x + \tan )^2}$

. . We have: . $\frac{dy}{dx} \;=\;\frac{-2\sec x(\sec x - \tan x)} {\sec x+ \tan x}$

$\text{Multiply by }\frac{\sec x + \tan x}{\sec x + \tan x}\!:$
. . . $\frac{dy}{dx} \;=\;\frac{-2\sec x(\sec x - \tan x)}{\sec x + \tan x}\cdot{\color{blue}\frac{\sec x + \tan x}{\sec x + \tan x}} \;=\; \frac{-2\sec x\overbrace{(\sec^2\!x - \tan^2\!x)}^{\text{This is 1}}}{(\sec x + \tan x)^2}$

Therefore: . $\frac{dy}{dx} \;=\;\frac{-2\sec x}{(\sec x + \tan x)^2}$