Question

find the derivative of the function:
(8 + sec x)/(8- sec x)

Answer 1

use the quotient rule :)

keep in mind that \(\bf \cfrac{d}{dx}[sec(x)]\implies sec(x)tan(x)\)

Answer 2

so, \[\sec x \tan x (8-\sec x ) - \sec x \tan x (8+ \sec x) / (8 -\sec x ) \]

Answer 3

(8 - sec x )^2 *

Answer 4

what do i do after? multiply whats inside the bracket?

Answer 5

hmmm one sec

Answer 6

\(\bf \cfrac{d}{dx}\left[\cfrac{(8+sec(x))}{(8-sec(x))}\right]
\\ \quad \\
\cfrac{[sec(x)tan(x)] (8-sec(x))\quad -\quad (8+sec(x))[-sec(x)tan(x)]}{(8-sec(x))^2}
\\ \quad \\
\cfrac{{\color{brown}{ [sec(x)tan(x)]}} (8-sec(x))\quad +\quad (8+sec(x)){\color{brown}{ [sec(x)tan(x)]}}}{(8-sec(x))^2}
\\ \quad \\
\cfrac{{\color{brown}{ [sec(x)tan(x)]}}\quad [(8\cancel{ -sec(x) })+(8\cancel{ +sec(x) })]}{(8-sec(x))^2}\)

Answer 7

so the 2 sec(x)'s above cancel out... so you're left with only 8+8 = 16 and the common factor in the numerator

Answer 8

\(\bf \cfrac{d}{dx}\left[\cfrac{(8+sec(x))}{(8-sec(x))}\right]
\\ \quad \\
\cfrac{[sec(x)tan(x)] (8-sec(x))\quad -\quad (8+sec(x))[-sec(x)tan(x)]}{(8-sec(x))^2}
\\ \quad \\
\cfrac{{\color{brown}{ [sec(x)tan(x)]}} (8-sec(x))\quad +\quad (8+sec(x)){\color{brown}{ [sec(x)tan(x)]}}}{(8-sec(x))^2}
\\ \quad \\
\cfrac{{\color{brown}{ [sec(x)tan(x)]}}\quad [(8\cancel{ -sec(x) })+(8\cancel{ +sec(x) })]}{(8-sec(x))^2}
\\ \quad \\
\cfrac{[sec(x)tan(x)][8+8]}{(8-sec(x))^2}\implies \cfrac{16[sec(x)tan(x)]}{(8-sec(x))^2}\)