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  1. #1
    Junior Member
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    showing

    hey pliz help me show that the differentiation of y=secx-tanx/(secx+tanx) is -2secx/(secx+tanx)^2
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Multiply top and bottom by sec(x) +tan(x) to obtain

    1/[sec(x)+tan(x)]^2

    f ' = -2 (sec(x)+tan(x))^(-3)* (sec(x)tan(x)+sec^2(x))

    now just clean it up
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  3. #3
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    Hello, Rose!

    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}

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