# Given that `y = x- ln (sec x + tan x)` , `0lt=xlt=pi/2` .Find dy/dx.

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You should differentiate the function with respect to x such that:

`(dy)/(dx) = 1 - 1/(sec x + tan x)*(sec x + tan x)'`

`(dy)/(dx) = 1 - 1/(sec x + tan x)*(1/cos x + sin x/cos x)'`

`(dy)/(dx) = 1 - 1/(sec x + tan x)*((1+sin x)/cos x)'`

Using the quotient rule yields:

`(dy)/(dx) = 1 - 1/(sec x + tan x)*((cos x*cos x +(1+sin x)sin x)/(cos^2 x))`

`(dy)/(dx) = 1 - 1/(sec x + tan x)*((cos^2 x + sin^2 x + 1)/(cos^2 x))`

Using the fundamental formula of trigonometry `cos^2 x + sin^2 x = ` 1 yields:

`(dy)/(dx) = 1 - 1/(sec x + tan x)*(2/(cos^2 x))`

`(dy)/(dx) = 1 - cos x/(1 + sin x)*(2/(cos^2 x))`

`(dy)/(dx) = 1 - 2/(cos x(1 + sin x))`

**Hence, evaluating the derivative of the given function yields `(dy)/(dx) = 1 - 2/(cos x(1 + sin x)).` **