We need to know the following trigonometric function here.
`cos2x = cos^x-sin^2x`
`sin2x = 2sinxcosx`
`sin^2x+cos^2x = 1`
`(1+tanx)/(1-tanx)`
`= (1+tanx)/(1-tanx)xx(1+tanx)/(1+tanx)`
`= (1+tanx)^2/(1-tan^2x)`
`= (1+sinx/cosx)^2/(1-(sin^2(x))/(cos^2(x)))`
`= ((cosx+sinx)^2/(cos^2x))/((cos^2x-sin^2x)/(cos^2x))`
`= (cosx+sinx)^2/(cos^2x-sin^2x)`
`= (cos^2x+sin^2x+2sinxcosx)/(cos2x)`
`= (1+sin2x)/(cos2x)`
`f(x) = (1+tanx)/(1-tanx) = (1+sin2x)/(cos2x)`
`f'(x)`
`= (cos2x*2cos2x-(1+sin2x)(-2sin2x))/(cos^2(2x))`
`= (2(cos^2(2x)+sin^2(2x)+sin2x))/(cos^2(2x))`
`= (2(1+sin2x))/(cos^2(2x))`
`= (2(1+sin2x))/(1-sin^2(2x))`
...
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We need to know the following trigonometric function here.
`cos2x = cos^x-sin^2x`
`sin2x = 2sinxcosx`
`sin^2x+cos^2x = 1`
`(1+tanx)/(1-tanx)`
`= (1+tanx)/(1-tanx)xx(1+tanx)/(1+tanx)`
`= (1+tanx)^2/(1-tan^2x)`
`= (1+sinx/cosx)^2/(1-(sin^2(x))/(cos^2(x)))`
`= ((cosx+sinx)^2/(cos^2x))/((cos^2x-sin^2x)/(cos^2x))`
`= (cosx+sinx)^2/(cos^2x-sin^2x)`
`= (cos^2x+sin^2x+2sinxcosx)/(cos2x)`
`= (1+sin2x)/(cos2x)`
`f(x) = (1+tanx)/(1-tanx) = (1+sin2x)/(cos2x)`
`f'(x)`
`= (cos2x*2cos2x-(1+sin2x)(-2sin2x))/(cos^2(2x))`
`= (2(cos^2(2x)+sin^2(2x)+sin2x))/(cos^2(2x))`
`= (2(1+sin2x))/(cos^2(2x))`
`= (2(1+sin2x))/(1-sin^2(2x))`
`= (2(1+sin2x))/((1+sin2x)(1-sin2x))`
`= 2/(1-sin2x)`
So the derivative of f(x) is `f'(x)= 2/(1-sin2x)`
You should use the quotient rule such that:
`f'(x) = ((1 + tan x)'(1 - tan x) + (1 + tan x)(1 - tan x)')/((1 - tan x)^2)`
`f'(x) = ((1/(cos^2 x))(1 - tan x) + (1 + tan x)(-1/(cos^2 x)))/((1 - tan x)^2)`
`f'(x) = (1/(cos^2 x) - tan x/(cos^2 x) - 1/(cos^2 x) -tan x/(cos^2 x))/((1 - tan x)^2)`
`f'(x) = (- 2tan x/(cos^2 x))/((1 - tan x)^2)`
Hence, evaluating the derivative of the function using the quotient rule yields `f'(x) = (- 2tan x/(cos^2 x))/((1 - tan x)^2).`