if y=1-tanx/1+tanx find dy?dx?

The answer is as follows:

-
Brackets MUST be used.
y = (1 - tan x) / (1 + tan x)

dy/dx is given by Quotient Rule :-

(1 + tan x) (- sec²x) - (1 - tan x)(sec²x)
---------------------------------------------------------------
(1 + tan x )²

1 - tan x sec²x - sec²x + tan x sec²x
-----------------------------------------------------------
(1 + tan x )²

1 - sec²x
---------------------
( 1 + tan x )²

- tan²x
-----------------------
( 1 + tan x )²

.......1 - tan(x)
y =---------------
.......1 + tan(x)

..............[1 + tan(x)] d/dx -sec^2(x) - [1 - tan(x)] d/dx sec^2(x)
dy/dx =--------------------------------------------------------------------------
............................[1 + tan(x)]^2

........- sec^2(x)[1 + tan(x)] - sec^2(x)][1 - tan(x)]
=--------------------------------------------------------------
.........................[1 + tan(x)]^2

......- sec^2(x) - sec^2(x)tan(x) - sec^2(x) + sec^2(x)tan(x)
= ----------------------------------------------------------------------------
........................[1 + tan(x)]^2

......- .2sec^2(x)
=---------------------- answer//
....[1 + tan(x)]^2

f(x) = [1 - tan(x)] / [1 + tan(x)] → you know that: tan(x) = sin(x)/cos(x)

f(x) = [1 - {sin(x)/cos(x)}] / [1 + {sin(x)/cos(x)}]

f(x) = [{cos(x) - sin(x)}/cos(x)] / [{cos(x) + sin(x)}/cos(x)]

f(x) = [cos(x) - sin(x)] / [cos(x) + sin(x)]


This function looks like (u/v), so its derivative looks like: [(u'.v) - (v'.u)]/v² → where:

u = cos(x) - sin(x) → u' = - sin(x) - cos(x)

v = cos(x) + sin(x) → v' = - sin(x) + cos(x)


f'(x) = [(u'.v) - (v'.u)]/v²

f'(x) = { [- sin(x) - cos(x)].[cos(x) + sin(x)] - [- sin(x) + cos(x)].[cos(x) - sin(x)] } / [cos(x) + sin(x)]²

f'(x) = { [- sin(x).cos(x) - sin²(x) - cos²(x) - cos(x).sin(x)] - [- sin(x).cos(x) + sin²(x) + cos²(x) - cos(x).sin(x)] } / [cos²(x) + 2.cos(x).sin(x) + sin²(x)]

f'(x) = [- sin(x).cos(x) - sin²(x) - cos²(x) - cos(x).sin(x) + sin(x).cos(x) - sin²(x) - cos²(x) + cos(x).sin(x)] / [cos²(x) + 2.cos(x).sin(x) + sin²(x)]

f'(x) = [- 2.sin²(x) - 2.cos²(x)] / [cos²(x) + 2.cos(x).sin(x) + sin²(x)]

f'(x) = - 2.[sin²(x) + cos²(x)] / [cos²(x) + 2.cos(x).sin(x) + sin²(x)] → recal: cos²(x) + sin²(x) = 1

f'(x) = - 2.[1] / [1 + 2.cos(x).sin(x)]

f'(x) = - 2/[1 + 2.cos(x).sin(x)] → recall the identity: sin(2x) = 2.sin(x).cos(x)

f'(x) = - 2/[1 + sin(2x)]

...but your result is right too.