Answer
$f'(x) = \dfrac{\sec x. \tan x + \sec x .\tan^2 x - \sec^3 x}{(1+\tan x)^2}$
Work Step by Step
In order to derivate this function, you have to apply the quotient rule
$\dfrac{d}{dx}(\dfrac{a}{b})=\dfrac{a'b-ab'}{b^2}$
So, let's identify a and b and derivate them
$a=\sec x$
$a'=\sec x. \tan x$
$b=1+\tan x$
$b'=\sec^2 x$
Then, substitute in the formula:
$f'(x) = \dfrac{(\sec x. \tan x)(1+\tan x)-(\sec x)(\sec^2 x)}{(1+\tan x)^2}$
Simplify and get the answer:
$f'(x) = \dfrac{\sec x. \tan x + \sec x .\tan^2 x - \sec^3 x}{(1+\tan x)^2}$
